Question

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?
y =
(x +

Answer 1

Answer:

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