Hi Bethany

If (-6,64) the coordinates that your parabola passes through?

Then your parabola is

f(x) = x^{2}/2 - 27x - 116

Based on your x intercepts

Your Parabola looks like

f(x) = a(x + 4)(x - 58)

since it passes through (-6,64) we can use that to find a

64 = a(-6+4)(-6-58)

64 =a(-2)(-64)

64 =a(128)

64/128 =a

1/2 =a

Multiplying gives the standard form

f(x) = x^{2}/2 -27x - 116 where

a = 1, b = -27 and c = -116

The x = h coordinate of the vertex is -b/2a = -(-27)/2(1/2) = 27/1 = 27

Plugging in 27 for x to get f(x) =y = k coordinate of the vertex

729/2 -729-116 = -480.5

vertex of your parabola is (27, -480.5)

Also vertex form of your parabola looks

f(x) = 1/2(x - 27)^{2} - 961/2

You can plot this at desmos.com to see what it looks like.

f(x) = x^{2}/2 - 27x - 116 passes through (-6, 64), (-4,0) and (58,0)

I do hope this helps

Brenda D.

10/23/20

Bethany S.

Personally, I didn't understand. It is from my math quiz.10/23/20

Katie S.

You have to find a first. You can not assume that a is 1. So if you bring 1/2 into your equation, it should be one that goes through all of those points. y= 1/2x^2 -27x -116. I just left it in intercept form bc it was not specific which form was needed. y=1/2(x+4)(x-58).10/23/20