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.

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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.

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

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!

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

What if it doesnt have an axis of symmetry?

we know that

If an axis of symmetry is [tex]x=-8[/tex] and has a maximum height of [tex]2[/tex]

then

Is a vertical parabola open down with vertex at [tex](-8,2)[/tex]

the equation in vertex form is equal to

[tex]y=a(x-h)^{2}+k[/tex]

where

(h,k) is the vertex

substitute

[tex]y=a(x+8)^{2}+2[/tex]

Find the value of a

the parabola passes through the point [tex](-7,-1)[/tex]

substitute in the formula

[tex]-1=a(-7+8)^{2}+2[/tex]

[tex]-1=a(1)^{2}+2[/tex]

[tex]-1=a+2[/tex]

[tex]a=-3[/tex]

therefore

the answer is

the equation in vertex form is [tex]y=-3(x+8)^{2}+2[/tex]