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

Answer 1

Answer:

The required equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$

Explanation:

Let us assume that the hole is at y = 0m, with x as the time.

From the question we have (-1s, 8m) as the vertex (here x being the time variable is supposed to be in seconds and y being the distance variable is supposed to be in meters)

At x = 1s, the ball gets to the hole, therefore we have point (1s, 0m)

We know that the vertex of the parabola y = ax² + bx + c is at

$x =\frac{-b}{2a}$

therefore we have:

$-1 = \frac{-b}{2a}$

We then have the following equations:

$8 = a\times -1^2 + b\times -1 + c$

$0 = a\times -1^2 + b\times 1 + c$

$-1 = \frac{-b}{2a}$

From the 3rd equation we have

1 X 2a = b.

Therefore we have:

$8 = a\times -1^2 - 1\times 2a\times1 + c$

$0 = a\times 1^2 + 1 \times2a\times 1 + c$

We can simplify both equations and get:

$8 = a\times( -1^2 - 2s^2) + c = -a\times 3^2 + c$

$0 = a\times(1^2 + 2^2) + c = a\times 3^2 + c$

The first equation now becomes:

$8 = -a\times 3 - a\times 3 = -a\times 6$

$a = frac{8}{-6} = -\frac{4}{3}$

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

$c = -a\times3 = -(-\frac{4}{3})*3 = 4$

$b = 1\times 2a = 1\times 2(-\frac{4}{3})= -\frac{8}{3}$

Then the equation is:

$y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4$