Question

Use these functions to answer the questions.

Answer 1

Part A
We have $f\left(x\right)=\left(x+3\right)^2-1$. To solve for the x-intercept, we set f(x) equal to 0.  That is
     $\left(x+3\right)^2-1=0$

     $\left(x+3\right)^2=1$

Take the square root of both sides, 
     $x+3=1$

     $x=-2$

The x-intercept is (-2,0). 

To solve for the y-intercept, we set x=0. That is 
     $y=\left(0+3\right)^2-1=3^2-1=9-1=8$

The y-intercept is (0, 8) 

The coordinates of the optimum point are actually the vertex which can be easily seen from the vertex form equation given above. The minimum point is (-3, -1).

Part B.
We have $g\left(x\right)=-2x^2+8x+3$.
Factor out -2
     $=-2\left(x^2-4x\right)+3$

Complete the square
     $=-2\left(x^2-4x+4\right)+3-2\left(4\right)$

Simplify
     $g(x)=-2\left(x-2\right)^2-5$

Part C
We have $h\left(t\right)=-16t^2+28t$.
The maximum height is 12.25 feet after 0.875 seconds from the time of the jump. The dolphin will be back in the water after 1.75 seconds. The graph of the jump is shown in the photo.