Find the x-intercepts of the parabola with vertex (-3,16) and y-intercept (0,-20).

Question

Find the x-intercepts of the parabola with vertex (-3,16) and y-intercept (0,-20).

Answer 1

Laurynn, I hope I can still help!

 

First, since we have a vertex, (h,k), we can plug in what we have, also using the other given point.

 

   y = a(x - h)² + k

-20 = a(0 -(-3))² + 16

-20 = 9a + 16

-36 = 9a

   a = -4

 

Now, we take our a of -4 and plug it in again. We know that y = 0 along the x-axis, but we need to find the x's to go with!

 

0 = -4(x - (-3))² + 16

-16 = -4(x - (-3))² 

4 = (x - (-3))²

2 = x + 3

x = -1

 

-2 = x + 3

x = -5

 

So, our x-intercepts are (-1,0) and (-5,0).

 

Bam.

 

 

Answer 2

We can write the function of the parabola in vertex form

 

y = a(x - h)² + k

 

where:

a is the coefficient of the x² term

the vertex is the coordinate (h, k)

 

y = a(x + 3)² + 16

 

 

Next, plug in the y-intercept into the vertex form to find a.  Set x=0 and y=-20.

 

-20 = a(0 + 3)² + 16

 

-20 = 9a + 16

 

-36 = 9a

 

-4 = a

 

 

The equation of the parabola is now

 

y = -4(x + 3)² + 16

 

 

To find the x-intercepts, we set y=0 and solve for x.

 

0 = -4(x + 3)² + 16

 

 

Subtract 16 on both sides of the equation.

 

-16 = -4(x + 3)²

 

 

Divide both sides of the equation by -4.

 

4 = (x + 3)²

 

 

Square-root both sides of the equation.

 

±√4 = x + 3

 

 

Subtract 3 on both sides of the equation.

 

-3 ± √4 = x

 

 

We end up with

 

x = -3 - √4               and              x = -3 + √4

 

 

Simplify both values of x to get your x-intercepts.