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Algebra 1

7.15.2 The Vertex Form

Algebra 17.15.2 The Vertex Form

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Activity

Here are two sets of equations for quadratic functions you saw earlier. In each set, the expressions that define the output are equivalent. In other words, all the equations in Set 1 define the same function. The same is true for Set 2: all the equations define the same function but they are just written in different forms.

Set 1:

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

g ( x ) = x ( x + 4 ) g ( x ) = x ( x + 4 )

h ( x ) = ( x + 2 ) 2 4 h ( x ) = ( x + 2 ) 2 4

Set 2:

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

q ( x ) = ( 5 x ) ( x 1 ) q ( x ) = ( 5 x ) ( x 1 )

r ( x ) = 1 ( x 3 ) 2 + 4 r ( x ) = 1 ( x 3 ) 2 + 4

The expression that defines h h is written in vertex form. We can show that it is equivalent to the expression defining f f by expanding and simplifying the expression:

( x + 2 ) 2 4 = ( x + 2 ) ( x + 2 ) 4 = x 2 + 2 x + 2 x + 4 4 = x 2 + 4 x , w h i c h i s f ( x ) ( x + 2 ) 2 4 = ( x + 2 ) ( x + 2 ) 4 = x 2 + 2 x + 2 x + 4 4 = x 2 + 4 x , w h i c h i s f ( x )

If this simplified expression is factored, then we arrive at the expression that defines function g g : x 2 + 4 x = x ( x + 4 ) x 2 + 4 x = x ( x + 4 ) . The expression that defines g g is written in factored form. The simplified expression that defines function f f is written in standard form.

1. Show that the expressions defining r r and p p are equivalent.

For questions 2 – 3, examine the graphs representing the quadratic functions.

...

2. Examine the graph of the function h h .

a. What is the x x -coordinate of the vertex of the graph of h h ?

b. What is the y y -coordinate of the vertex of the graph of h h ?

3. Examine the graph of the function r r .

...

a. What is the x x -coordinate of the vertex of the graph of r r ?

b. What is the y y -coordinate of the vertex of the graph of r r ?

4. Why do you think expressions such as those defining h h and r r are said to be written in vertex form?

Why Should I Care?

A hand holds a tablet displaying a diagram of projectile motion: a cannon fires a ball at angle theta, following a curved path; the horizontal distance the cannonball travels is represented by s and labeled at the bottom of the trajectory. The height of the cannonball from the ground is represented by the variable h and labeled with a vertical line under the path.

When you launch an object, it does not travel in a straight line. Instead, it follows a curved path called a parabola. This means you have to account for distance, height, and angle if you want to hit your target. You can use quadratic equations to find these measurements.

The next time you throw a ball, think about the parabolic path that it follows.

Self Check

Which equation is equivalent to y = ( x 3 ) 2 9 ?
  1. y = x 2 + 6 x 9
  2. y = x 2 6 x
  3. y = x 2 3 x 9
  4. y = x 2 18

Additional Resources

Equivalent Forms of Quadratics

Quadratics can be written in different equivalent forms:

  • Standard form: y = a x 2 + b x + c y = a x 2 + b x + c
  • Factored form: y = ( x + a ) ( x + b ) y = ( x + a ) ( x + b )
  • Vertex form: y = a ( x h ) 2 + k y = a ( x h ) 2 + k

Example

Write y = ( x + 2 ) 2 9 y = ( x + 2 ) 2 9 in standard and factored forms.

Here is how to find the equivalent standard form:

Step 1 - Expand the vertex form to multiply the binomials.

y = ( x + 2 ) ( x + 2 ) 9 y = ( x + 2 ) ( x + 2 ) 9

Step 2 - Multiply the binomials.

y = x 2 + 2 x + 2 x + 4 9 y = x 2 + 2 x + 2 x + 4 9

Step 3 - Combine like terms.

y = x 2 + 4 x 5 y = x 2 + 4 x 5

Here is how to find the factored form:

Step 1 - From the standard form, find the factors of the trinomial expression y = x 2 + 4 x 5 y = x 2 + 4 x 5 .

Step 2 - Examine the trinomial expression.

x 2 + 4 x 5 x 2 + 4 x 5

Step 3 - Factor the trinomial.

( x + 5 ) ( x 1 ) ( x + 5 ) ( x 1 )

The factored form is y = ( x + 5 ) ( x 1 ) y = ( x + 5 ) ( x 1 ) .

Try it

Try It: Equivalent Forms of Quadratics

Find the equation in standard form that is equivalent to y = ( x 2 ) 2 + 4 y = ( x 2 ) 2 + 4 .

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