Answer:
Given equation that shows the height in feet after x seconds,
[tex]f(x) = -3x^2 - 6x + 21[/tex]
[tex]f(x) = -3(x^2 + 2x) + 21[/tex]
[tex]f(x) = -3(x^2 + 2x + 1) + 21 + 3[/tex]
[tex]f(x) = -3(x+1)^2+24[/tex]
Since, the vertex form of a quadratic equation is,
[tex]f(x) = a(x-h)^2 + k[/tex]
Where,
(h, k) is the vertex of the equation.
By comparing,
Vertex = (h, k) = (-1, 24)
i.e. h = -1
Now, f(0) = 21
⇒ The initial height of the rock from the lake is 21 ft,
Now, for zeroes,
f(x) = 0,
[tex]-3x^2 - 6x + 21=0[/tex]
By quadratic formula,
[tex]x = \frac{6 \pm \sqrt{(-6)^2 - 4\times -3\times 21}}{-6}[/tex]
[tex]=\frac{6\pm \sqrt{36 + 252}}{-6}[/tex]
[tex]\implies x = 1.828\text{ or }x = -3.828[/tex]
Since, number of seconds can not be zero,
∴ x = 1.828 seconds
i.e. the rock will reach in the lack after 1.828 seconds.
