Solution
1.
We need to write y = x^2 -2x - 5 in vertex so as to be able to determine the vertex and the axis of symmetry.
To do that, we need to write in the form y = a(x - h)^2 + k -----(1)
[tex]\begin{gathered} y=x^2-2x-5=x^2-2x+(-\frac{2}{2})^2-(-\frac{2}{2})^2-5 \\ \\ \Rightarrow y=x^2-2x+1-1-5 \\ \\ \Rightarrow y=x^2-2x+1-6 \\ \\ \Rightarrow y=(x-1)^2-6\text{ -----\lparen2\rparen} \end{gathered}[/tex]
Comparing equation (2) with equation (1)
here a = 1, h = 1, k = -6.
Therefore, the vertex is: (1, -6)
The axis of symmetry is: x = h
Therefore, the axis of symmetry is x = 1.
Question 2)
The height function is given as;
[tex]H(t)=-16t^2+16t+32[/tex]
To find out how long Jason will hit the water, we need to set H(t) = 0;
[tex]\begin{gathered} \Rightarrow-16t^2+16t+32=0 \\ \\ \Rightarrow-16(t^2-t-2)=0 \\ \\ \Rightarrow t^2-t-2=0 \\ \\ \Rightarrow(t-2)(t+1)=0 \\ \\ \Rightarrow t=2,t=-1 \end{gathered}[/tex]
since negative time is not lucid, t = 2 is cogent.
Therefore, it will take 2 seconds for Jason to hit the water.
