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nika2105 [10]
3 years ago
6

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 reach

es 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
Mathematics
1 answer:
sukhopar [10]3 years ago
6 0

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

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