Respuesta :
Answer:
The required equation is:
[tex]y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4[/tex]
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
[tex]x =\frac{-b}{2a}[/tex]
therefore we have:
[tex]-1 = \frac{-b}{2a}[/tex]
We then have the following equations:
[tex]8 = a\times -1^2 + b\times -1 + c[/tex]
[tex]0 = a\times -1^2 + b\times 1 + c[/tex]
[tex]-1 = \frac{-b}{2a}[/tex]
From the 3rd equation we have
1 X 2a = b.
Therefore we have:
[tex]8 = a\times -1^2 - 1\times 2a\times1 + c[/tex]
[tex]0 = a\times 1^2 + 1 \times2a\times 1 + c[/tex]
We can simplify both equations and get:
[tex]8 = a\times( -1^2 - 2s^2) + c = -a\times 3^2 + c[/tex]
[tex]0 = a\times(1^2 + 2^2) + c = a\times 3^2 + c[/tex]
The first equation now becomes:
[tex]8 = -a\times 3 - a\times 3 = -a\times 6[/tex]
[tex]a = frac{8}{-6} = -\frac{4}{3}[/tex]
With a, we can find the values of c and b.
[tex]c = -a\times3 = -(-\frac{4}{3})*3 = 4 [/tex]
[tex]b = 1\times 2a = 1\times 2(-\frac{4}{3})= -\frac{8}{3}[/tex]
Then the equation is:
[tex]y = -\frac{4}{3}t^2 -\frac{8}{3}t + 4[/tex]
