# Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write

Suppose a parabola has an axis of symmetry at x = -8 , a maximum height of 2, and passes through the point (-7, -1). write the equation of the parabola in vertex form.

## This Post Has 5 Comments

1. torrejes241 says:

As it has a maximum value the coefficient of x^2 will be negative

The vertex will be at (-8,2)  so in vertex form it is

y = a(x + 8)^2 + 2
and as it passes through (-7,-1) we have:

-1  = a(-7+8)^2 + 2

-1 = a + 2  so a = -3

answer is  y = -3(x + 8)^2 + 2

in standard form this is y = -3x^2 -48x  - 190

2. alisonlebron15 says:

Its standard form of equation: (x-h)^2=-4p (y-k), (h,k)=(x,y) coordinates of the vertex
For given parabola: Vertex = (-8, 2)
(x-(-8))² = -4p(y - 2)

solve for 4p using coordinates of given point ( -7, -1)
-7+8  = 4p (-1 - 2)
1 = 4p(-3)
4p = -1/3

So, Equation of given parabola:
(x+8)² = -(y-2) / 3

Hope this helps!

3. genyjoannerubiera says:

Y=-3(x+8)^2+2 Fill in the vertex (-8,2), and the ordered pair(-7.1) then solve for a. Y = a(x-h)^2 + k

4. melanie1055 says:

What if it doesnt have an axis of symmetry?

5. itsbrizee says:

we know that

If an axis of symmetry is $x=-8$ and has a maximum height of $2$

then

Is a vertical parabola open down with vertex at $(-8,2)$

the equation in vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex

substitute

$y=a(x+8)^{2}+2$

Find the value of a

the parabola passes through the point $(-7,-1)$

substitute in the formula

$-1=a(-7+8)^{2}+2$

$-1=a(1)^{2}+2$

$-1=a+2$

$a=-3$

therefore

the equation in vertex form is  $y=-3(x+8)^{2}+2$