The Linear Term in a Quadratic Expression

7.13.2 The Linear Term in a Quadratic Expression

7.13.2 • The Linear Term in a Quadratic Expression

Activity

1. Use the graphing tool or technology outside the course. Graph $y = x^{2}$ , and then experiment with adding different linear terms (for example, $x^{2} + 4 x$ , $x^{2} + 20 x$ , $x^{2} - 50 x$ ). Record your observations.

Access multimedia content

Compare your answer:

The graph still looks like a quadratic function when adding a linear term but it "shifts" both horizontally and vertically. Adding a positive multiple of $x$ to $x^{2}$ shifts it down and to the left while adding a negative multiple of $x$ shifts it down and to the right.

Use your observations from question 1 to answer questions 2 – 5 and complete the table.

equation	$x$ -intercepts	$x$ -coordinate of vertex
$y = x^{2} + 6 x$	2a. _____	2b. _____
$y = x^{2} - 10 x$	3a. _____	3b. _____
$y = - x^{2} + 50 x$	4a. _____	4b. _____
$y = - x^{2} - 36 x$	5a. _____	5b. _____

2. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = x^{2} + 6 x$ .

a. What are the $x$ -intercepts for the equation $y = x^{2} + 6 x$ ?

Compare your answer:

$(0, 0)$ , $(- 6, 0)$

b. What is the $x$ -coordinate of the vertex for the equation $y = x^{2} + 6 x$ ?

-3

3. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = x^{2} - 10 x$ .

a. What are the $x$ -intercepts for the equation $y = x^{2} - 10 x$ ?

Compare your answer:

$(0, 0)$ , $(10, 0)$

b. What is the $x$ -coordinate of the vertex for the equation $y = x^{2} - 10 x$ ?

4. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = - x^{2} + 50 x$ .

a. What are the $x$ -intercepts for the equation $y = - x^{2} + 50 x$ ?

Compare your answer:

$(0, 0)$ , $(50, 0)$

b. What is the $x$ -coordinate of the vertex for the equation $y = - x^{2} + 50 x$ ?

5. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = - x^{2} - 36 x$ .

a. Find the $x$ -intercepts for the equation $y = - x^{2} - 36 x$ .

Compare your answer:

$(0, 0)$ , $(- 36, 0)$

b. What is the $x$ -coordinate of the vertex for the equation $y = - x^{2} - 36 x$ ?

-18

Some quadratic expressions have no linear terms. For questions 6 – 7, find the $x$ -intercepts and the $x$ -coordinate of the vertex of the graph representing each equation:

$y = x^{2} - 25$

$y = x^{2} + 16$

(Note it is possible for the graph to not intersect the $x$ -axis.) If you get stuck, try graphing the equations.

6. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = x^{2} - 25$ .

a. What are the $x$ -intercepts for the equation $y = x^{2} - 25$ ?

Compare your answer:

$(5, 0)$ , $(- 5, 0)$

b. What is the $x$ -coordinate of the vertex for the equation $y = x^{2} - 25$ ?

7. Find the $x$ -intercepts and the $x$ -coordinate of the vertex for the equation $y = x^{2} + 16$ .

a. What are the $x$ -intercepts for the equation $y = x^{2} + 16$ ?

Compare your answer:

There are no $x$ -intercepts.

b. What is the $x$ -coordinate of the vertex for the equation $y = x^{2} + 16$ ?

Self Check

Find the

x

-coordinates of all of the

x

-intercepts of the function

f(x)=x^2-5x

$x = 0$ , $x = -5$
$x = 0$
$x = 0$ , $x = 5$
$x = 5$

Additional Resources

Quadratics Without Linear Terms

Example

Find the $x$ -intercepts and the vertex for $f (x) = x^{2} - 4$ .

Step 1 - Write the function in factored form. Use difference of squares:

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

Step 2 - Set each factor equal to 0.

$x - 2 = 0$

$x = 2$

$x + 2 = 0$

$x = - 2$

The $x$ -intercepts are $(2, 0)$ and $(- 2, 0)$ .

Step 3 - Find the $x$ -value in the middle of the two $x$ -intercepts. This occurs at $x = 0$ . This is the $x$ -coordinate of the vertex.

To find the point at the vertex, substitute into the function:

$f (0) = 0^{2} - 4 = - 4$

Vertex: $(0, - 4)$

Try it

Quadratics Without Linear Terms

Find the $x$ -intercepts and the vertex for $f (x) = x^{2} - 64$ .

Here is how to find the $x$ -intercepts and vertex of $f (x) = x^{2} - 64$ :

Step 1 - Write the function in factored form:
$f (x) = x^{2} - 64 = (x - 8) (x + 8)$

Step 2 - Set each factor equal to 0.
$x - 8 = 0$
$x = 8$

$x + 8 = 0$
$x = - 8$

The $x$ -intercepts are at $(8, 0)$ and $(- 8, 0)$ .

The vertex is located halfway between the $x$ -intercepts, at $x = 0$ .

Step 3 - To find the point, substitute into $f (x) = x^{2} - 64$ .

$f (0) = 0^{2} - 64 = - 64$

The vertex is at $(0, - 64)$ .