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Algebra 1

7.13.2 The Linear Term in a Quadratic Expression

Algebra 17.13.2 The Linear Term in a Quadratic Expression

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Activity

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

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

equation x x -intercepts x x -coordinate of vertex

y = x 2 + 6 x y = x 2 + 6 x

2a. _____

2b. _____

y = x 2 10 x y = x 2 10 x

3a. _____

3b. _____

y = x 2 + 50 x y = x 2 + 50 x

4a. _____

4b. _____

y = x 2 36 x y = x 2 36 x

5a. _____

5b. _____

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

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

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

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

a. What are the x x -intercepts for the equation y = x 2 10 x y = x 2 10 x ?

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

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

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

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

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

a. Find the x x -intercepts for the equation y = x 2 36 x y = x 2 36 x .

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

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

y = x 2 25 y = x 2 25

y = x 2 + 16 y = x 2 + 16

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

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

a. What are the x x -intercepts for the equation y = x 2 25 y = x 2 25 ?

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

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

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

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

Self Check

Find the x -coordinates of all of the x -intercepts of the function f ( x ) = x 2 5 x .
  1. x = 0 , x = 5
  2. x = 0
  3. x = 0 , x = 5
  4. x = 5

Additional Resources

Quadratics Without Linear Terms

Example

Find the x x -intercepts and the vertex for f ( x ) = x 2 4 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 ) f ( x ) = x 2 4 = ( x 2 ) ( x + 2 )

Step 2 - Set each factor equal to 0.

x 2 = 0 x 2 = 0

x = 2 x = 2

x + 2 = 0 x + 2 = 0

x = 2 x = 2

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

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

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

f ( 0 ) = 0 2 4 = 4 f ( 0 ) = 0 2 4 = 4

Vertex: ( 0 , 4 ) ( 0 , 4 )

Try it

Try It: Quadratics Without Linear Terms

Find the x x -intercepts and the vertex for f ( x ) = x 2 64 f ( x ) = x 2 64 .

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