Answer:
see the explanation
Step-by-step explanation:
we have the quadratic equation
[tex]y=-16x^{2} +32x-10[/tex]
This is a vertical parabola open downward
The vertex is a maximum
Find the x-intercepts of the quadratic equation
The x-intercepts are the values of x when the value of y is equal to zero
so
For y=0
[tex]-16x^{2} +32x-10=0[/tex]
Solve the quadratic equation
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]-16x^{2} +32x-10=0[/tex]
so
[tex]a=-16\\b=32\\c=-10[/tex]
substitute in the formula
[tex]x=\frac{-32(+/-)\sqrt{32^{2}-4(-16)(-10)}} {2(-16)}[/tex]
[tex]x=\frac{-32(+/-)\sqrt{384}} {-32}[/tex]
[tex]x=\frac{-32(+/-)8\sqrt{6}}{-32}[/tex]
[tex]x=\frac{-32(+)8\sqrt{6}}{-32}=\frac{32(-)8\sqrt{6}}{32}=0.39\ sec[/tex]
[tex]x=\frac{-32(-)8\sqrt{6}}{-32}=\frac{32(+)8\sqrt{6}}{32}=1.61\ sec[/tex]
therefore
This parabola has two x-intercepts representing the times when the dolphin's height above water is zero feet
