30pts PLEASE HELP I WILL GIVE BRAINLIEST TO RIGHT ANSWER The distance of a golf ball from the hole can be represented by the right side of a parabola
with vertex (-1,8). The ball reaches the hole 1 second after it is hit, at time 0.

What is the equation of the parabola, in vertex form, that represents the ball's distance
from the hole, y, at time x?

Let's suppose that the hole is at y = 0m, where x is the time variable.

we know that:

The vertex is (-1s, 8m).

(i suppose x in seconds and y in meters)

At x = 1s, the ball reaches the hole, so we also have the point:

(1s, 0m).

Remember that the vertex of a quadratic equation y = a*x^2 + b*x + c is at:

x = -b/2a,

then we have:

-1 = -b/2a.

Then we have tree equations:

8m = a*(-1s)^2 + b*-1s + c

0m = a*(1s)^2 + b*1s + c

-1s = -b/2a.

First we should isolate one variable in the third equation, and then replace it in one of the other two:

1s*2a = b.

So we can replace b in the first two equations bi 1s*2a.

8m = a*1s^2 - 1s*2a*1s + c

0m = a*1s^2 + 1s*2a*1s + c

We can simplify both equations and get:

8m = a*( 1s^2 - 2s^2) + c = -a*1s^2 + c.

0m = a*(1s^2 + 2s^2) + c = a*3s^2 + c.

Easily we can isolate c in the second equation and then replace it into the first equation:

c = -a*3s^2

The first equation becomes:

8m = -a*1s^2 - a*3s^2 = -a*4s^2

a = 8m/-4s^2 = -2m/s^2.

Now with a, we can find the values of c and b.

c = -a*3s^2 = -(-2m/s^2)*3s^2 = 6m.

b = 1s*2a = 1s*(-2m/s^2) = -2m/s.

Then the equation is:

y = (-2m/s^2)*t^2 + (-2m/s)*t + 6m

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