Graph Quadratic Functions Using Transformations

9.7 Graph Quadratic Functions Using Transformations

Learning Objectives

By the end of this section, you will be able to:

  • Graph quadratic equations of the form f(x)=x2+kf(x)=x2+k
  • Graph quadratic functions of the form f(x)=(x-h)2f(x)=(x-h)2
  • Graph quadratic functions of the form f(x)=ax2f(x)=ax2
  • Graph quadratic functions using transformations
  • Find a quadratic function from its graph

Be Prepared 9.19

Before you get started, take this readiness quiz.

Graph the function f(x)=x2f(x)=x2 by plotting points.
If you missed this problem, review Example 3.54.

Be Prepared 9.20

Factor completely: y2−14y+49.y2−14y+49.
If you missed this problem, review Example 6.24.

Be Prepared 9.21

Factor completely: 2x2−16x+32.2x2−16x+32.
If you missed this problem, review Example 6.26.

Graph Quadratic Functions of the form f(x)=x2+kf(x)=x2+k

In the last section, we learned how to graph quadratic functions using their properties. Another method involves starting with the basic graph of f(x)=x2f(x)=x2 and ‘moving’ it according to information given in the function equation. We call this graphing quadratic functions using transformations.

In the first example, we will graph the quadratic function f(x)=x2f(x)=x2 by plotting points. Then we will see what effect adding a constant, k, to the equation will have on the graph of the new function f(x)=x2+k.f(x)=x2+k.

Example 9.53

Graph f(x)=x2,g(x)=x2+2,f(x)=x2,g(x)=x2+2, and h(x)=x2−2h(x)=x2−2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

Solution

Plotting points will help us see the effect of the constants on the basic f(x)=x2f(x)=x2 graph. We fill in the chart for all three functions.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals x squared plus 2, the ordered pair (x, g of x), h of x equals x squared minus 2, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals x squared plus 2 column contains the expressions 9 plus 2, 4 plus 2, 1 plus 2, 0 plus 2, 1 plus 2, 4 plus 2, and 9 plus 2. The (x, g of x) column has the ordered pairs of (negative 3, 11), (negative 2, 6), (negative 1, 3), (0, 2), (1, 3), (2, 6), and (3, 11). In the h of x equals x squared minus 2 column, the expressions given are 9 minus 2, 4 minus 2, 1 minus 2, 0 minus 2, 1 minus 2, 4 minus 2, and 9 minus 2. In last column, (x, h of x), contains the ordered pairs (negative 3, 7), (negative 2, 2), (negative 1, negative 1), (0, negative 2), (1, negative 1), (2, 2), and (3, 7).

The g(x) values are two more than the f(x) values. Also, the h(x) values are two less than the f(x) values. Now we will graph all three functions on the same rectangular coordinate system.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle is the graph of f of x equals x squared has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The top parabola has been moved up 2 units, and the bottom has been moved down 2 units.

The graph of g(x)=x2+2g(x)=x2+2 is the same as the graph of f(x)=x2f(x)=x2 but shifted up 2 units.

The graph of h(x)=x2−2h(x)=x2−2 is the same as the graph of f(x)=x2f(x)=x2 but shifted down 2 units.

Try It 9.105

ⓐ Graph f(x)=x2,g(x)=x2+1,f(x)=x2,g(x)=x2+1, and h(x)=x2−1h(x)=x2−1 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Try It 9.106

ⓐ Graph f(x)=x2,g(x)=x2+6,f(x)=x2,g(x)=x2+6, and h(x)=x2−6h(x)=x2−6 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

The last example shows us that to graph a quadratic function of the form f(x)=x2+k,f(x)=x2+k, we take the basic parabola graph of f(x)=x2f(x)=x2 and vertically shift it up (k>0)(k>0) or shift it down (k<0)(k<0).

This transformation is called a vertical shift.

Graph a Quadratic Function of the form f ( x ) = x 2 + k f ( x ) = x 2 + k Using a Vertical Shift

The graph of f(x)=x2+kf(x)=x2+k shifts the graph of f(x)=x2f(x)=x2 vertically k units.

  • If k > 0, shift the parabola vertically up k units.
  • If k < 0, shift the parabola vertically down |k||k| units.

Now that we have seen the effect of the constant, k, it is easy to graph functions of the form f(x)=x2+k.f(x)=x2+k. We just start with the basic parabola of f(x)=x2f(x)=x2 and then shift it up or down.

It may be helpful to practice sketching f(x)=x2f(x)=x2 quickly. We know the values and can sketch the graph from there.

This figure shows an upward-opening parabola on the x y-coordinate plane, with vertex (0, 0). Other points on the curve are located at (negative 4, 16), (negative 3, 9), (negative 2, 4), (negative 1, 1), (1, 1), (2, 4), (3, 9), and (4, 16).

Once we know this parabola, it will be easy to apply the transformations. The next example will require a vertical shift.

Example 9.54

Graph f(x)=x2−3f(x)=x2−3 using a vertical shift.

Solution

We first draw the graph of f(x)=x2f(x)=x2 on
the grid.
This figure shows an upward-opening parabola on the x y-coordinate plane with a vertex of (0, 0) with other points on the curve located at (negative 1, 1) and (1, 1). It is the graph of f of x equals x squared.
Determine kk. Two mathematical equations are displayed: f(x) = x^2 + k and f(x) = x^2 - 3 (highlighted in red). The image appears to be for a math problem determining the value of k.
The mathematical equation k = -3 is displayed in black text on a plain white background, centrally positioned within the frame.
Shift the graph f(x)=x2f(x)=x2 down 3. This figure shows 2 upward-opening parabolas on the x y-coordinate plane. The top curve is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The bottom curve has been moved down 3 units.
This table illustrates steps and their visual outcomes for graphing transformations of functions, specifically showing shifts of f(x) = x^2.

Try It 9.107

Graph f(x)=x2−5f(x)=x2−5 using a vertical shift.

Try It 9.108

Graph f(x)=x2+7f(x)=x2+7 using a vertical shift.

Graph Quadratic Functions of the form f(x)=(x−h)2f(x)=(x−h)2

In the first example, we graphed the quadratic function f(x)=x2f(x)=x2 by plotting points and then saw the effect of adding a constant k to the function had on the resulting graph of the new function f(x)=x2+k.f(x)=x2+k.

We will now explore the effect of subtracting a constant, h, from x has on the resulting graph of the new function f(x)=(x−h)2.f(x)=(x−h)2.

Example 9.55

Graph f(x)=x2,g(x)=(x−1)2,f(x)=x2,g(x)=(x−1)2, and h(x)=(x+1)2h(x)=(x+1)2 on the same rectangular coordinate system. Describe what effect adding a constant to the function has on the basic parabola.

Solution

Plotting points will help us see the effect of the constants on the basic f(x)=x2f(x)=x2 graph. We fill in the chart for all three functions.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals the quantity of x minus 1 squared, the ordered pair (x, g of x), h of x equals the quantity of x plus 1 squared, and the ordered pair (x, h of x). In the x column, the values given are negative 3, negative 2, negative 1, 0, 1, 2, and 3. In the f of x equals x squared column, the values are 9, 4, 1, 0, 1, 4, and 9. In the (x, f of x) column, the ordered pairs (negative 3, 9), (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), (2, 4), and (3, 9) are given. The g of x equals the quantity of x minus 1 squared column contains the values of 16, 9, 4, 1, 0, 1, and 4. The (x, g of x) column has the ordered pairs of (negative 3, 1), (negative 2, 9), (negative 1, 4), (0, 1), (1, 0), (2, 1), and (3, 4). In the h of x equals the quantity of x plus 1 squared, the values given are 4, 1, 0, 1, 4, 9, and 16. In last column, (x, h of x), contains the ordered pairs (negative 3, 4), (negative 2, 1), (negative 1, 0), (0, 4), (1, negative 1), (2, 9), and (3, 16).

The g(x) values and the h(x) values share the common numbers 0, 1, 4, 9, and 16, but are shifted.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. The middle curve is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The left curve has been moved to the left 1 unit, and the right curve has been moved to the right 1 unit. The figure says on the first line that the graph of g of x equals the quantity x minus 1 square is the same as the graph of f of x equals x squared but shifted right 1 unit. The second line states that the graph of h of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The third line of the figure says g of x equals the quantity x minus 1 squared with an arrow underneath it pointing to the right with 1 unit written beside it. Finally, it gives h of x equals the quantity of x plus 1 squared with an arrow underneath it pointing to the left with 1 unit written beside it.

Try It 9.109

ⓐ Graph f(x)=x2,g(x)=(x+2)2,f(x)=x2,g(x)=(x+2)2, and h(x)=(x−2)2h(x)=(x−2)2 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

Try It 9.110

ⓐ Graph f(x)=x2,g(x)=x2+5,f(x)=x2,g(x)=x2+5, and h(x)=x2−5h(x)=x2−5 on the same rectangular coordinate system.
ⓑ Describe what effect adding a constant to the function has on the basic parabola.

The last example shows us that to graph a quadratic function of the form f(x)=(x−h)2,f(x)=(x−h)2, we take the basic parabola graph of f(x)=x2f(x)=x2 and shift it left (h > 0) or shift it right (h < 0).

This transformation is called a horizontal shift.

Graph a Quadratic Function of the form f ( x ) = ( x − h ) 2 f ( x ) = ( x − h ) 2 Using a Horizontal Shift

The graph of f(x)=(x−h)2f(x)=(x−h)2 shifts the graph of f(x)=x2f(x)=x2 horizontally hh units.

  • If h > 0, shift the parabola horizontally right h units.
  • If h < 0, shift the parabola horizontally left |h||h| units.

Now that we have seen the effect of the constant, h, it is easy to graph functions of the form f(x)=(x−h)2.f(x)=(x−h)2. We just start with the basic parabola of f(x)=x2f(x)=x2 and then shift it left or right.

The next example will require a horizontal shift.

Example 9.56

Graph f(x)=(x−5)2f(x)=(x−5)2 using a horizontal shift.

Solution

We first draw the graph of f(x)=x2f(x)=x2 on
the grid.
A coordinate plane displays the graph of a parabola y=x^2. The curve opens upwards, symmetric about the y-axis, with its vertex located at the origin (0,0).
Determine h. Two mathematical equations are shown: f(x) = (x - h) squared in black and f(x) = (x - 5) squared in red, demonstrating a quadratic function and its specific form.
The image displays the equation 'h = 5' in a light gray font against a clean white background.
Shift the graph f(x)=x2f(x)=x2 to the right 5 units. Graph of y=x^2 (blue) and y=(x-5)^2 (red), demonstrating a horizontal translation of 5 units to the right on the coordinate plane.

Try It 9.111

Graph f(x)=(x−4)2f(x)=(x−4)2 using a horizontal shift.

Try It 9.112

Graph f(x)=(x+6)2f(x)=(x+6)2 using a horizontal shift.

Now that we know the effect of the constants h and k, we will graph a quadratic function of the form f(x)=(x−h)2+kf(x)=(x−h)2+k by first drawing the basic parabola and then making a horizontal shift followed by a vertical shift. We could do the vertical shift followed by the horizontal shift, but most students prefer the horizontal shift followed by the vertical.

Example 9.57

Graph f(x)=(x+1)2−2f(x)=(x+1)2−2 using transformations.

Solution

This function will involve two transformations and we need a plan.

Let’s first identify the constants h, k.

F of x equals the quantity x plush 1 squared minus 2 is given on the top line with f of x equals the quanitity x minus h squared minis k on the second line. The given equation was changed to f of x equals the quantity of x minus negative 1 squared plush negative 2 on the third line. The final line says h equals negative 1 and k equals negative 2.

The h constant gives us a horizontal shift and the k gives us a vertical shift.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared with an arrow coming from it pointing to f of x equals the quantity x plus 1 squared minus 2. The next lines say h equals negative 1 which means shift left 1 unit and k equals negative 2 which means shift down 2 units.

We first draw the graph of f(x)=x2f(x)=x2 on the grid.

The figure says on the first line that the graph of f of x equals the quantity x plus 1 squared is the same as the graph of f of x equals x squared but shifted left 1 unit. The second line states that the graph of f of x equals the quantity x plus 1 squared minus 2 is the same as the graph of f of x equals the quantity x plus 1 squared but shifted down 2 units. The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 1, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left one unit to produce f of x equals the quantity of x plus 1 squared. By moving f of x equals the quantity of x plus 1 squared down 1, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 1, then another curve moved down 1 to produce f of x equals the quantity of x plus 1 squared minus 2.

Try It 9.113

Graph f(x)=(x+2)2−3f(x)=(x+2)2−3 using transformations.

Try It 9.114

Graph f(x)=(x−3)2+1f(x)=(x−3)2+1 using transformations.

Graph Quadratic Functions of the Form f(x)=ax2f(x)=ax2

So far we graphed the quadratic function f(x)=x2f(x)=x2 and then saw the effect of including a constant h or k in the equation had on the resulting graph of the new function. We will now explore the effect of the coefficient a on the resulting graph of the new function f(x)=ax2.f(x)=ax2.

A table depicting the effect of constants on the basic function of x squared. The table has seven columns labeled x, f of x equals x squared, the ordered pair (x, f of x), g of x equals 2 times x squared, the ordered pair (x, g of x), h of x equals one-half times x squared, and the ordered pair (x, h of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared column, the values are 4, 1, 0, 1, and 4. In the (x, f of x) column, the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 2 times x squared column contains the expressions 2 times 4, 2 times 1, 2 times 0, 2 times 1, and 2 times 4. The (x, g of x) column has the ordered pairs of (negative 2, 8), (negative 1, 2), (0, 0), (1, 2), and (2,8). In the h of x equals one-half times x squared, the expressions given are one-half times 4, one-half times 1, one-half times 0, one-half times 1, and one-half times 4. In last column, (x, h of x), contains the ordered pairs (negative 2, 2), (negative 1, one-half), (0, 0), (1, one-half), and (2, 2).

If we graph these functions, we can see the effect of the constant a, assuming a > 0.

This figure shows 3 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). The slimmer curve of g of x equals 2 times x square has a vertex at (0,0) and other points of (negative 1, one-half) and (1, one-half). The wider curve, h of x equals one-half x squared, has a vertex at (0,0) and other points of (negative 2, 2) and (2,2).

To graph a function with constant a it is easiest to choose a few points on f(x)=x2f(x)=x2 and multiply the y-values by a.

Graph of a Quadratic Function of the form f ( x ) = a x 2 f ( x ) = a x 2

The coefficient a in the function f(x)=ax2f(x)=ax2 affects the graph of f(x)=x2f(x)=x2 by stretching or compressing it.

  • If 0<|a|<1,0<|a|<1, the graph of f(x)=ax2f(x)=ax2 will be “wider” than the graph of f(x)=x2.f(x)=x2.
  • If |a|>1|a|>1, the graph of f(x)=ax2f(x)=ax2 will be “skinnier” than the graph of f(x)=x2.f(x)=x2.

Example 9.58

Graph f(x)=3x2.f(x)=3x2.

Solution

We will graph the functions f(x)=x2f(x)=x2 and g(x)=3x2g(x)=3x2 on the same grid. We will choose a few points on f(x)=x2f(x)=x2 and then multiply the y-values by 3 to get the points for g(x)=3x2.g(x)=3x2.

The table depicts the effect of constants on the basic function of x squared. The table has 3 columns labeled x, f of x equals x squared with the ordered pair (x, f of x), and g of x equals 3 times x squared with the ordered pair (x, g of x). In the x column, the values given are negative 2, negative 1, 0, 1, and 2. In the f of x equals x squared with the ordered pair (x, f of x), the ordered pairs (negative 2, 4), (negative 1, 1), (0, 0), (1, 1), and (2, 4) are given. The g of x equals 3 times x squared with the ordered pair (x, g of x) column has the ordered pairs of (negative 2, 12) because 3 times 4 equals 12, (negative 1, 3) because 3 times 1 equals 3, (0, 0) because 3 times 0 equals 0, (1, 3) because 3 times 1 equals 3, and (2,12) because 3 times 4 equals 12. The graph beside the table shows 2 upward-opening parabolas on the x y-coordinate plane. One is the graph of f of x equals x squared and has a vertex of (0, 0). Other points given on the curve are located at (negative 2, 4) (negative 1, 1), (1, 1), and (2,4). The slimmer curve of g of x equals 3 times x squared has a vertex at (0,0) and other points given of (negative 2, 12), (negative 1, 3), (1, 3), and (2,12).

Try It 9.115

Graph f(x)=−3x2.f(x)=−3x2.

Try It 9.116

Graph f(x)=2x2.f(x)=2x2.

Graph Quadratic Functions Using Transformations

We have learned how the constants a, h, and k in the functions, f(x)=x2+k,f(x)=(x−h)2,f(x)=x2+k,f(x)=(x−h)2, and f(x)=ax2f(x)=ax2 affect their graphs. We can now put this together and graph quadratic functions f(x)=ax2+bx+cf(x)=ax2+bx+c by first putting them into the form f(x)=a(x−h)2+kf(x)=a(x−h)2+k by completing the square. This form is sometimes known as the vertex form or standard form.

We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations.

This figure shows the difference when completing the square with a quadratic equation and a quadratic function. For the quadratic equation, start with x squared plus 8 times x plus 6 equals zero. Subtract 6 from both sides to get x squared plus 8 times x equals negative 6 while leaving space to complete the square. Then, complete the square by adding 16 to both sides to get x squared plush 8 times x plush 16 equals negative 6 plush 16. Factor to get the quantity x plus 4 squared equals 10. For the quadratic function, start with f of x equals x squared plus 8 times x plus 6. The second line shows to leave space between the 8 times x and the 6 in order to complete the square. Complete the square by adding 16 and subtracting 16 on the same side to get f of x equals x squared plus 8 times x plush 16 plus 6 minus 16. Factor to get f of x equals the quantity of x plush 4 squared minus 10.

When we complete the square in a function with a coefficient of x2 that is not one, we have to factor that coefficient from just the x-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the x-terms.

Once we get the constant we want to complete the square, we must remember to multiply it by that coefficient before we then subtract it.

Example 9.59

Rewrite f(x)=−3x2−6x−1f(x)=−3x2−6x−1 in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k form by completing the square.

Solution

A mathematical equation is presented, showing the function f(x) equal to -3x^2 - 6x - 1. This quadratic equation is displayed in a clear, digital font on a white background.
Separate the x terms from the constant. A mathematical equation is displayed on a white background, reading 'f(x) = -3x^2 - 6x - 1' in black text.
Factor the coefficient of x2x2, −3−3. The image displays the quadratic function f(x) = -3(x^2 + 2x) - 1.
Prepare to complete the square. The equation f(x) = -3(x^2 + 2x) - 1 is shown, with a blank space after '2x' indicating an incomplete expression, possibly for completing the square.
Take half of 2 and then square it to complete the
square. (12·2)2=1(12·2)2=1
The constant 1 completes the square in the
parentheses, but the parentheses is multiplied by
−3−3. So we are really adding −3−3 We must then
add 3 to not change the value of the function.
A mathematical equation f(x) = -3(x^2 + 2x + 1) - 1 + 3. It demonstrates how adding 1 inside the parenthesis, multiplied by -3, means you must add 3 outside to balance the expression.
Rewrite the trinomial as a square and subtract the
constants.
A mathematical equation is displayed against a white background. The equation reads: f(x) = -3(x + 1)^2 + 2. It represents a quadratic function in vertex form.
The function is now in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k
form.
Two mathematical equations are displayed: the vertex form of a quadratic function, f(x) = a(x - h) ^2 + k, followed by a specific example, f(x) = -3(x + 1) ^2 + 2.

Try It 9.117

Rewrite f(x)=−4x2−8x+1f(x)=−4x2−8x+1 in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k form by completing the square.

Try It 9.118

Rewrite f(x)=2x2−8x+3f(x)=2x2−8x+3 in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k form by completing the square.

Once we put the function into the f(x)=(x−h)2+kf(x)=(x−h)2+k form, we can then use the transformations as we did in the last few problems. The next example will show us how to do this.

Example 9.60

Graph f(x)=x2+6x+5f(x)=x2+6x+5 by using transformations.

Solution

Step 1. Rewrite the function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k vertex form by completing the square.

A mathematical equation is displayed, showing a quadratic function: f(x) = x^2 + 6x + 5. The function is written in a standard algebraic notation with variables and coefficients.
Separate the x terms from the constant. A mathematical equation is displayed, showing a quadratic function: f(x) = x^2 + 6x + 5. The equation is rendered in black text on a white background, appearing as a standard algebraic expression.
Take half of 6 and then square it to complete the square.
(12·6)2=9(12·6)2=9
We both add 9 and subtract 9 to not change the value of the function. A mathematical equation for f(x) is shown as f(x) = x^2 + 6x + 9 + 5 - 9, with the numbers +9 and -9 highlighted in red.
Rewrite the trinomial as a square and subtract the constants. A mathematical equation is displayed on a white background: f(x) = (x + 3)^2 - 4. This represents a quadratic function in vertex form.
The function is now in the f(x)=(x−h)2+kf(x)=(x−h)2+k form. Two mathematical equations are displayed: the general vertex form of a parabola, f(x) = (x - h)^2 + k, in red, and a specific instance, f(x) = (x + 3)^2 - 4, in black.

Step 2: Graph the function using transformations.

Looking at the h, k values, we see the graph will take the graph of f(x)=x2f(x)=x2 and shift it to the left 3 units and down 4 units.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared with an arrow coming from it pointing to f of x equals the quantity x plus 3 squared minus 4. The next lines say h equals negative 3 which means shift left 3 unit and k equals negative 4 which means shift down 4 units

We first draw the graph of f(x)=x2f(x)=x2 on the grid.

To graph f of x equals the quantity x plus 3 squared, shift the graph of f of x equals x squares to the left 3 units. To graph f of x equals the quantity x plus 3 squared minus 4, shift the graph the quantity x plus 3 squared down 4 units. The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By shifting that graph of f of x equals x squared left 3, we move to the next graph, which shows the original f of x equals x squared and then another curve moved left 3 units to produce f of x equals the quantity of x plus 3 squared. By moving f of x equals the quantity of x plus 3 squared down 2, we move to the final graph, which shows the original f of x equals x squared and the f of x equals the quantity of x plus 3 squared, then another curve moved down 4 to produce f of x equals the quantity of x plus 1 squared minus 4.

Try It 9.119

Graph f(x)=x2+2x−3f(x)=x2+2x−3 by using transformations.

Try It 9.120

Graph f(x)=x2−8x+12f(x)=x2−8x+12 by using transformations.

We list the steps to take to graph a quadratic function using transformations here.

How To

Graph a quadratic function using transformations.

  1. Step 1. Rewrite the function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form by completing the square.
  2. Step 2. Graph the function using transformations.

Example 9.61

Graph f(x)=−2x2−4x+2f(x)=−2x2−4x+2 by using transformations.

Solution

Step 1. Rewrite the function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k vertex form by completing the square.

The quadratic function f(x) = -2x^2 - 4x + 2 is displayed on a white background.
Separate the x terms from the constant. A mathematical equation for a quadratic function is displayed, reading f(x) = -2x^2 - 4x + 2.
We need the coefficient of x2x2 to be one.
We factor −2−2 from the x-terms.
A mathematical equation is displayed, showing a function f(x) = -2(x^2 + 2x) + 2. This is a quadratic function written in a form that can be simplified or used for completing the square.
Take half of 2 and then square it to complete the square.
(12·2)2=1(12·2)2=1
We add 1 to complete the square in the parentheses, but the parentheses is multiplied by −2−2. Se we are really adding −2−2. To not change the value of the function we add 2. An algebraic expression f(x) = -2(x^2 + 2x + 1) + 2 + 2, highlighting the addition of 1 inside the parenthesis and the compensating addition of 2 outside, as shown by the red numbers and arrow.
Rewrite the trinomial as a square and subtract the constants. The image displays the equation of a quadratic function in vertex form: f(x) = -2(x + 1)^2 + 4.
The function is now in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k form. Two equations for quadratic functions, with the general vertex form f(x) = a(x - h)^2 + k shown in red, and a specific example f(x) = -2(x + 1)^2 + 4 in black.

Step 2. Graph the function using transformations.

F of x equals x squared is given with an arrow coming from it pointing to f of x equals negative 2 times x squared with an arrow coming from it pointing to f of x equals negative 2 times the quantity x plus 1 squared. An arrow come from it to point to f of x equals negative 2 times the quantity x plus 1 squared plus 4. The next line says a equals negative 2 which means multiply the y-values by negative 2, then h equals negative 1 which means shift left 1 unit and k equals 4 which means shift up 4 units

We first draw the graph of f(x)=x2f(x)=x2 on the grid.

To graph f of x equals negative 2 times x squared, multiply the y-values in parabola of f of x equals x squared by negative 2. To graph f of x equals negative 2 times the quantity x plus 1 squared, shift the graph of f of x equals negative 2 times x squared to the left 1 unit. To graph f of x equals negative 2 times the quantity x plus 1 squared plus 4, shift the graph of f of x equals negative 2 times the quantity x plus 1 squared up 4 units. The first graph shows 1 upward-opening parabola on the x y-coordinate plane. It is the graph of f of x equals x squared which has a vertex of (0, 0). Other points on the curve are located at (negative 1, 1) and (1, 1). By multiplying by negative 2, move to the next graph showing the original f of x equals x squared and the new slimmer and flipped graph of f of x equals negative 2 x squared. By shifting that graph of f of x equals negative 2 times x squared left 1, we move to the next graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and then another curve moved left 1 unit to produce f of x equals negative 2 times the quantity of x plus 1 squared. By moving f of x equals negative 2 times the quantity of x plus 1 squared up 4, we move to the final graph, which shows the original f of x equals x squared, f of x equals negative 2 x squared, and the f of x equals negative 2 times the quantity of x plus 1 squared, then another curve moved up 4 to produce f of x equals negative 2 times the quantity of x plus 1 squared plus 4.

Try It 9.121

Graph f(x)=−3x2+12x−4f(x)=−3x2+12x−4 by using transformations.

Try It 9.122

Graph f(x)=−2x2+12x−9f(x)=−2x2+12x−9 by using transformations.

Now that we have completed the square to put a quadratic function into f(x)=a(x−h)2+kf(x)=a(x−h)2+k form, we can also use this technique to graph the function using its properties as in the previous section.

If we look back at the last few examples, we see that the vertex is related to the constants h and k.

The first graph shows an upward-opening parabola on the x y-coordinate plane with a vertex of (negative 3, negative 4) with other points of (0, negative 5) and (0, negative 1). Underneath the graph, it shows the standard form of a parabola, f of x equals the quantity x minus h squared plus k, with the equation of the parabola f of x equals the quantity of x plus 3 squared minus 4 where h equals negative 3 and k equals negative 4. The second graph shows a downward-opening parabola on the x y-coordinate plane with a vertex of (negative 1, 4) and other points of (0,2) and (negative 2,2). Underneath the graph, it shows the standard form of a parabola, f of x equals a times the quantity x minus h squared plus k, with the equation of the parabola f of x equals negative 2 times the quantity of x plus 1 squared plus 4 where h equals negative 1 and k equals 4.

In each case, the vertex is (h, k). Also the axis of symmetry is the line x = h.

We rewrite our steps for graphing a quadratic function using properties for when the function is in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form.

How To

Graph a quadratic function in the form f(x)=a(x−h)2+kf(x)=a(x−h)2+k using properties.

  1. Step 1. Rewrite the function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form.
  2. Step 2. Determine whether the parabola opens upward, a > 0, or downward, a < 0.
  3. Step 3. Find the axis of symmetry, x = h.
  4. Step 4. Find the vertex, (h, k).
  5. Step 5. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
  6. Step 6. Find the x-intercepts.
  7. Step 7. Graph the parabola.

Example 9.62

ⓐ Rewrite f(x)=2x2+4x+5f(x)=2x2+4x+5 in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Solution

Rewrite the function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k
form by completing the square.
f(x)=2x2+4x+5f(x)=2x2+4x+5
f(x)=2(x2+2x)+5f(x)=2(x2+2x)+5
f(x)=2(x2+2x+1)+5−2f(x)=2(x2+2x+1)+5−2
f(x)=2(x+1)2+3f(x)=2(x+1)2+3
Identify the constants a,h,k.a,h,k. a=2h=−1k=3a=2h=−1k=3
Since a=2a=2, the parabola opens upward. A red U-shaped arrow with arrowheads pointing upwards and outwards at both ends, indicating a curve, cycle, or expansion.
The axis of symmetry is x=hx=h. The axis of symmetry is x=−1x=−1.
The vertex is (h,k)(h,k). The vertex is (−1,3)(−1,3).
Find the y-intercept by finding f(0)f(0). f(0)=2⋅02+4⋅0+5f(0)=2⋅02+4⋅0+5
f(0)=5f(0)=5
y-intercept (0,5)(0,5)
Find the point symmetric to (0,5)(0,5) across the
axis of symmetry.
(−2,5)(−2,5)
Find the x-intercepts. The discriminant negative, so there are
no x-intercepts. Graph the parabola.
A parabola graphed on a coordinate plane. The vertex is at (-1, 3), and points (-2, 5) and (0, 5) are marked. A dashed line at x=-1 represents the axis of symmetry.

Try It 9.123

ⓐ Rewrite f(x)=3x2−6x+5f(x)=3x2−6x+5 in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Try It 9.124

ⓐ Rewrite f(x)=−2x2+8x−7f(x)=−2x2+8x−7 in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form and ⓑ graph the function using properties.

Find a Quadratic Function from its Graph

So far we have started with a function and then found its graph.

Now we are going to reverse the process. Starting with the graph, we will find the function.

Example 9.63

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 2, negative 1) and y-intercept (0, 7).

Solution

Since it is quadratic, we start with the f(x)=a(x−h)2+kform.f(x)=a(x−h)2+kform.
The vertex, (h,k),(h,k), is (−2,−1)(−2,−1) so h=−2h=−2 and k=−1.k=−1. f(x)=a(x−(−2))2−1f(x)=a(x−(−2))2−1
To find aa, we use the yy-intercept, (0,7)(0,7).
So f(0)=7f(0)=7. 7=a(0+2)2−17=a(0+2)2−1
Solve for aa. 7=4a−17=4a−1
8=4a8=4a
2=a2=a
Write the function. f(x)=a(x−h)2+kf(x)=a(x−h)2+k
Substitute in h=−2,k=−1h=−2,k=−1 and a=2a=2. f(x)=2(x+2)2−1f(x)=2(x+2)2−1
Steps for deriving the equation of a quadratic function in vertex form, given its vertex and y-intercept.

Try It 9.125

Write the quadratic function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form whose graph is shown.

The graph shown is an upward facing parabola with vertex (3, negative 4) and y-intercept (0, 5).

Try It 9.126

Determine the quadratic function whose graph is shown.

The graph shown is an upward facing parabola with vertex (negative 3, negative 1) and y-intercept (0, 8).

Media

Section 9.7 Exercises

Practice Makes Perfect

Graph Quadratic Functions of the form f(x)=x2+kf(x)=x2+k

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, k, to the function has on the basic parabola.

293.

f(x)=x2,g(x)=x2+4,f(x)=x2,g(x)=x2+4, and h(x)=x2−4.h(x)=x2−4.

294.

f(x)=x2,g(x)=x2+7,f(x)=x2,g(x)=x2+7, and h(x)=x2−7.h(x)=x2−7.

In the following exercises, graph each function using a vertical shift.

295.

f ( x ) = x 2 + 3 f ( x ) = x 2 + 3

296.

f ( x ) = x 2 − 7 f ( x ) = x 2 − 7

297.

g ( x ) = x 2 + 2 g ( x ) = x 2 + 2

298.

g ( x ) = x 2 + 5 g ( x ) = x 2 + 5

299.

h ( x ) = x 2 − 4 h ( x ) = x 2 − 4

300.

h ( x ) = x 2 − 5 h ( x ) = x 2 − 5

Graph Quadratic Functions of the form f(x)=(x−h)2f(x)=(x−h)2

In the following exercises, ⓐ graph the quadratic functions on the same rectangular coordinate system and ⓑ describe what effect adding a constant, hh, inside the parentheses has

301.

f(x)=x2,g(x)=(x−3)2,f(x)=x2,g(x)=(x−3)2, and h(x)=(x+3)2.h(x)=(x+3)2.

302.

f(x)=x2,g(x)=(x+4)2,f(x)=x2,g(x)=(x+4)2, and h(x)=(x−4)2.h(x)=(x−4)2.

In the following exercises, graph each function using a horizontal shift.

303.

f ( x ) = ( x − 2 ) 2 f ( x ) = ( x − 2 ) 2

304.

f ( x ) = ( x − 1 ) 2 f ( x ) = ( x − 1 ) 2

305.

f ( x ) = ( x + 5 ) 2 f ( x ) = ( x + 5 ) 2

306.

f ( x ) = ( x + 3 ) 2 f ( x ) = ( x + 3 ) 2

307.

f ( x ) = ( x − 5 ) 2 f ( x ) = ( x − 5 ) 2

308.

f ( x ) = ( x + 2 ) 2 f ( x ) = ( x + 2 ) 2

In the following exercises, graph each function using transformations.

309.

f ( x ) = ( x + 2 ) 2 + 1 f ( x ) = ( x + 2 ) 2 + 1

310.

f ( x ) = ( x + 4 ) 2 + 2 f ( x ) = ( x + 4 ) 2 + 2

311.

f ( x ) = ( x − 1 ) 2 + 5 f ( x ) = ( x − 1 ) 2 + 5

312.

f ( x ) = ( x − 3 ) 2 + 4 f ( x ) = ( x − 3 ) 2 + 4

313.

f ( x ) = ( x + 3 ) 2 − 1 f ( x ) = ( x + 3 ) 2 − 1

314.

f ( x ) = ( x + 5 ) 2 − 2 f ( x ) = ( x + 5 ) 2 − 2

315.

f ( x ) = ( x − 4 ) 2 − 3 f ( x ) = ( x − 4 ) 2 − 3

316.

f ( x ) = ( x − 6 ) 2 − 2 f ( x ) = ( x − 6 ) 2 − 2

Graph Quadratic Functions of the form f(x)=ax2f(x)=ax2

In the following exercises, graph each function.

317.

f ( x ) = −2 x 2 f ( x ) = −2 x 2

318.

f ( x ) = 4 x 2 f ( x ) = 4 x 2

319.

f ( x ) = −4 x 2 f ( x ) = −4 x 2

320.

f ( x ) = − x 2 f ( x ) = − x 2

321.

f ( x ) = 1 2 x 2 f ( x ) = 1 2 x 2

322.

f ( x ) = 1 3 x 2 f ( x ) = 1 3 x 2

323.

f ( x ) = 1 4 x 2 f ( x ) = 1 4 x 2

324.

f ( x ) = − 1 2 x 2 f ( x ) = − 1 2 x 2

Graph Quadratic Functions Using Transformations

In the following exercises, rewrite each function in the f(x)=a(x−h)2+kf(x)=a(x−h)2+k form by completing the square.

325.

f ( x ) = −3 x 2 − 12 x − 5 f ( x ) = −3 x 2 − 12 x − 5

326.

f ( x ) = 2 x 2 − 12 x + 7 f ( x ) = 2 x 2 − 12 x + 7

327.

f ( x ) = 3 x 2 + 6 x − 1 f ( x ) = 3 x 2 + 6 x − 1

328.

f ( x ) = −4 x 2 − 16 x − 9 f ( x ) = −4 x 2 − 16 x − 9

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form and ⓑ graph it by using transformations.

329.

f ( x ) = x 2 + 6 x + 5 f ( x ) = x 2 + 6 x + 5

330.

f ( x ) = x 2 + 4 x − 12 f ( x ) = x 2 + 4 x − 12

331.

f ( x ) = x 2 + 4 x + 3 f ( x ) = x 2 + 4 x + 3

332.

f ( x ) = x 2 − 6 x + 8 f ( x ) = x 2 − 6 x + 8

333.

f ( x ) = x 2 − 6 x + 15 f ( x ) = x 2 − 6 x + 15

334.

f ( x ) = x 2 + 8 x + 10 f ( x ) = x 2 + 8 x + 10

335.

f ( x ) = − x 2 + 8 x − 16 f ( x ) = − x 2 + 8 x − 16

336.

f ( x ) = − x 2 + 2 x − 7 f ( x ) = − x 2 + 2 x − 7

337.

f ( x ) = − x 2 − 4 x + 2 f ( x ) = − x 2 − 4 x + 2

338.

f ( x ) = − x 2 + 4 x − 5 f ( x ) = − x 2 + 4 x − 5

339.

f ( x ) = 5 x 2 − 10 x + 8 f ( x ) = 5 x 2 − 10 x + 8

340.

f ( x ) = 3 x 2 + 18 x + 20 f ( x ) = 3 x 2 + 18 x + 20

341.

f ( x ) = 2 x 2 − 4 x + 1 f ( x ) = 2 x 2 − 4 x + 1

342.

f ( x ) = 3 x 2 − 6 x − 1 f ( x ) = 3 x 2 − 6 x − 1

343.

f ( x ) = −2 x 2 + 8 x − 10 f ( x ) = −2 x 2 + 8 x − 10

344.

f ( x ) = −3 x 2 + 6 x + 1 f ( x ) = −3 x 2 + 6 x + 1

In the following exercises, ⓐ rewrite each function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form and ⓑ graph it using properties.

345.

f ( x ) = 2 x 2 + 4 x + 6 f ( x ) = 2 x 2 + 4 x + 6

346.

f ( x ) = 3 x 2 − 12 x + 7 f ( x ) = 3 x 2 − 12 x + 7

347.

f ( x ) = − x 2 + 2 x − 4 f ( x ) = − x 2 + 2 x − 4

348.

f ( x ) = −2 x 2 − 4 x − 5 f ( x ) = −2 x 2 − 4 x − 5

Matching

In the following exercises, match the graphs to one of the following functions: ⓐ f(x)=x2+4f(x)=x2+4 ⓑ f(x)=x2−4f(x)=x2−4 ⓒ f(x)=(x+4)2f(x)=(x+4)2 ⓓ f(x)=(x−4)2f(x)=(x−4)2 ⓔ f(x)=(x+4)2−4f(x)=(x+4)2−4 ⓕ f(x)=(x+4)2+4f(x)=(x+4)2+4 ⓖ f(x)=(x−4)2−4f(x)=(x−4)2−4 ⓗ f(x)=(x−4)2+4f(x)=(x−4)2+4

349.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, 0) and other points (negative 4, 4) and (negative 2, 4).
350.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (0, negative 4) and other points (negative 2, 0) and (2, 0).
351.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, negative 4) and other points (negative 4, 0) and (negative 2, 0).
352.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 4, 4) and other points (negative 6, 8) and (negative 2, 8).
353.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, 0) and other points (2, 4) and (2, 4).
354.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (0, 4) and other points (negative 2, 8) and (2, 8).
355.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, negative 4) and other points (2,0) and (6,0).
356.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (4, 4) and other points (2,8) and (6,8).

Find a Quadratic Function from its Graph

In the following exercises, write the quadratic function in f(x)=a(x−h)2+kf(x)=a(x−h)2+k form whose graph is shown.

357.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 5) and y-intercept (0, negative 4).
358.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (2,4) and y-intercept (0, 8).
359.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (1, negative 3) and y-intercept (0, negative 1).
360.
This figure shows an upward-opening parabola on the x y-coordinate plane. It has a vertex of (negative 1, negative 5) and y-intercept (0, negative 3).

Writing Exercise

361.

Graph the quadratic function f(x)=x2+4x+5f(x)=x2+4x+5 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

362.

Graph the quadratic function f(x)=2x2−4x−3f(x)=2x2−4x−3 first using the properties as we did in the last section and then graph it using transformations. Which method do you prefer? Why?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This figure is a list to assess your understanding of the concepts presented in this section. It has 4 columns labeled I can…, Confidently, With some help, and No-I don’t get it! Below I can…, there is graph Quadratic Functions of the form f of x equals x squared plus k; graph Quadratic Functions of the form f of x equals the quantity x minus h squared; graph Quadratic functions of the form f of x equals a times x squared; graph Quadratic Functions Using Transformations; find a Quadratic Function from its Graph. The other columns are left blank for you to check you understanding.

ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?