Graph Functions in Vertex Form

7.16.2 Graph Functions in Vertex Form

7.16.2 • Graph Functions in Vertex Form

Activity

Examine these two equations that define quadratic functions.

$p (x) = - (x - 4)^{2} + 10$

$q (x) = \frac{1}{2} (x - 4)^{2} + 10$

1. The graph of $p$ passes through $(0, - 6)$ and $(4, 10)$ , as shown on the coordinate plane.

Graph of two points on a coordinate plane. The points graphed and labeled include (0, negative 6) and (4, 10).

a. Find the coordinates of another point on the graph of $p$ . Explain or show your reasoning.

Compare your answer:

$(8, - 6)$ . If we find $p (8)$ and substitute 8 for $x$ in the expression, we have: $p (8) = - (8 - 4)^{2} + 10$ , which is equal to $- 16 + 10$ or -6.

b. Use the graphing tool or technology outside the course. Plot $p (x) = - (x - 4)^{2} + 10$ and the points $(0, - 6)$ , $(4, 10)$ , and $(8, - 6)$ to label the graph.

Access multimedia content

Compare your answer:

Graph of a parabola on a coordinate plane. The parabola passes through two identified and labeled points: (0, negative 6) and (4, 10). A third point that is on the parabola and symmetric to (0, negative 6) is marked but not labeled.

The graph of a quadratic function is symmetrical across the vertical line containing the vertex. If a point on the graph that is 4 units to the left of the line containing the vertex has a $y$ -value of -6, there is a point to the right of the line containing the vertex that also has a $y$ -value of -6.

A line containing the vertex is called the parabola’s axis of symmetry.

2. Find the $x$ - and $y$ -coordinates of the vertex and sketch the graph.

a. What is the $x$ -coordinate of the vertex of $q (x) = \frac{1}{2} (x - 4)^{2} + 10$ ?

4. The $x$ -coordinate of the vertex is found by using the squared term of the vertex form. The $x$ -coordinate of the vertex of the function $q$ is 4.

b. What is the the $y$ -coordinate of the vertex of $q (x) = \frac{1}{2} (x - 4)^{2} + 10$ ?

10. In vertex form, the $y$ -coordinate is the constant term. The value of the $y$ -coordinate of the vertex of the function $q$ is 10.

c. Use the graphing tool or technology outside the course. On the same coordinate plane containing the points and equation from question 1, plot the coordinates of the vertex of the function $q$ .

Compare your answer:

Graphs of two parabolas on a coordinate plane. The green parabola that opens down passes through three identified and labeled points: (0, negative 6), (4, 10), and (8, negative 6). The red parabola also passes through (4, 10) but opens up.

d. Find two other points that are on the graph of $q$ . Try to identify points that are symmetrical over the axis of symmetry.

Compare your answer:

One possible answer could be $(2, 12)$ and $(6, 12)$ . Substituting 2 for $x$ , we have: $q (2) = \frac{1}{2} (2 - 4)^{2} + 10$ , which is 12, so $(2, 12)$ is one point on the graph. Substituting 6 for $x$ , we have: $q (6) = \frac{1}{2} 1 (6 - 4)^{2} + 10$ , which equals 12, so $(6, 12)$ is another point.

Another possible answer could be $(0, 18)$ and $(8, 18)$ . Substituting 0 for $x$ , we get $q (0) = 18$ , so $(0, 18)$ is one point on the graph. This point is 4 units to the left and 8 units up from the vertex. Because the graph has a vertical line of symmetry, or axis of symmetry, through the vertex, it also goes through a point that is 4 units to the right and 8 units up, which is $(8, 18)$ .

Access multimedia content

e. Use the graphing tool or technology outside the course. Sketch and label the graph of $q$ . Explain or show your reasoning.

Compare your answer:

3. Recall that $p (x) = - (x - 4)^{2} + 10$ . Pilar says, "Once I know the vertex is $(4, 10)$ , I can find out, without graphing, whether the vertex is the maximum or the minimum of function $p$ . I would just compare the coordinates of the vertex with the coordinates of a point on either side of it."

Complete the table by answering questions a and b.

$x$	$3$	$4$	$5$
$p (x)$		$10$

Partial probability distribution fragment for variable x, showing p(4)=10.

a. Evaluate $p (3)$ .

9. $p (3) = - (3 - 4)^{2} + 10 = - (- 1)^{2} + 10 = - 1 + 10 = 9$ .

b. Evaluate $p (5)$ .

9. $p (5) = - (5 - 4)^{2} + 10 = - (1)^{2} + 10 = - 1 + 10 = 9$ .

c. Explain how Pilar might have reasoned whether the vertex is the minimum or maximum.

Compare your answer:

The values of $p (3)$ and $p (5)$ are both less than 10. The vertex of a quadratic graph represents the minimum or the maximum of the function. Because the two points on either side of the vertex of $p$ have a lesser $y$ -value, the vertex must be the maximum.

Are you ready for more?

Extending Your Thinking

Choose the equation for a quadratic function whose graph has the vertex at $(2, 3)$ and contains the point $(0, - 5)$ .

$y = - 2 (x - 2)^{2} + 3$
$y = - 3 (x - 2)^{2} + 2$
$y = - 5 (x - 2)^{2}$
$y = 2 (x - 2)^{2} + 3$

$y = - 2 {(x - 2)}^{2} + 3$

Use the graphing tool or technology outside the course. Sketch a graph of your function.

Compare your answer:

Graph of a parabola on a coordinate plane. The x-axis has a scale of 1 extending from negative 8 to 8. The vertical-axis extends from negative 14 to 14 with a scale of 2.

Self Check

The equation of a quadratic function is

y=(x-1)^2+5

and the point

(3,9)

also lies on the parabola. What is another point that lies on the parabola?

$(-1, 5)$
$(-1,9)$
$(3, 1)$
$(-5, 9)$

Additional Resources

Using Key Points to Graph Quadratics

We saw that vertex form is especially helpful for finding the vertex of a graph of a quadratic function. For example, we can tell that the function $p$ given by $p (x) = (x - 3)^{2} + 1$ has a vertex at $(3, 1)$ .

Graph of a parabola on a coordinate plane. The x-axis has a scale of 1 extending from 0 to 6. The y-axis extends from 0 to 10 with a scale of 2. The point (3, 1) on the parabola is labeled.

We also noticed that, when the squared expression $(x - 3)^{2}$ has a positive coefficient, the graph opens upward. You can use the vertex $(3, 1)$ to determine that minimum function value is 1.

But why does the function $p$ take on its minimum value when $x$ is 3?

Here is one way to explain it: When $x = 3$ , the squared term $(x - 3)^{2}$ equals 0, as $(3 - 3)^{2} = 0^{2} = 0$ . When $x$ is any other value besides 3, the squared term $(x - 3)^{2}$ is a positive number greater than 0. (Squaring any number results in a positive number.) This means that the output when $x \neq 3$ will always be greater than the output when $x = 3$ , so the function $p$ has a minimum value at $x = 3$ .

This table shows some values of the function for some values of $x$ . Notice that the output is the least when $x = 3$ and it increases both as $x$ increases and as it decreases.

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$(x - 3)^{2} + 1$	$10$	$5$	$2$	$1$	$2$	$5$	$10$

This table displays selected x-values and their corresponding function outputs for (x-3)^2 + 1.

The squared term sometimes has a negative coefficient, for instance: $h (x) = - 2 (x + 4)^{2}$ . The $x$ value that makes $(x + 4)^{2}$ equal 0 is -4, because $(- 4 + 4)^{2} = 0^{2} = 0$ . Any other $x$ value makes $(x + 4)^{2}$ greater than 0. But when $(x + 4)^{2}$ is multiplied by a negative number (-2), the resulting expression, $- 2 (x + 4)^{2}$ , ends up being negative. This means that the output when $x - 4$ will always be less than the output when $x = - 4$ , so the function $h$ has its maximum value when $x = - 4$ .

Graph of a parabola on a coordinate plane. The x-axis has a scale of 1 extending from negative 7 to 0. The y-axis extends from negative 8 to 2 with a scale of 2. The point (negative 4, 0) on the parabola is labeled.

Remember that we can find the $y$ -intercept of the graph representing any function we have seen. The $y$ -coordinate of the $y$ -intercept is the value of the function when $x = 0$ . If $g$ is defined by $g (x) = (x + 1)^{2} - 5$ , then the $y$ -intercept is $(0, - 4)$ because $g (0) = (0 + 1)^{2} - 5 = - 4$ . Its vertex is at $(- 1, - 5)$ . Another point on the graph with the same $y$ -coordinate is located the same horizontal distance from the vertex but on the other side.

Graph of a parabola on a coordinate plane. The x-axis has a scale of 1 extending from negative 4 to 2. The y-axis extends from negative 8 to 2 with a scale of 2. The points (negative 2, negative 4), (negative 1, negative 5), and (0, negative 4) have been labeled on the parabola.

Try it

Using Key Points to Graph Quadratics

A quadratic function is represented by the equation $y = (x + 3)^{2} - 5$ . If $(5, 61)$ is another point on the graph, find a third point on the curve.

Here is how to find another point on the curve:

One way to find another point on the curve is to use symmetry.

Since the vertex is $(- 3, - 5)$ and one point on the curve is $(5, 61)$ , there is another point on the other side of the vertex. Keep in mind, the $y$ -coordinate will remain at 61. To find the $x$ -coordinate, we need to determine how far apart horizontally the vertex and the other point are located from each other. There are 8 units from -3 to 5, so the new point will also be 8 units from -3 but to the left. Thus, the $x$ -coordinate will be at -11. So, there is another point at $(- 11, 61)$ .