Question

The football coach threw a football from a platform to his

Answer 1

Answer:

1)27.25 feet

2)-0.2774,2.027

3)Domain = All real numbers

Range = y∈R : $y \leq \frac{109}{4}$

Step-by-step explanation:

The height of the football, h, at time t seconds is modeled by the equation $h(t) = -16t^2 + 28t + 15$

General quadratic equation : $ax^2=bx+c=0$

1) Maximum height will be at $\frac{-b}{2a}=\frac{-28}{2(-16)}=\frac{7}{8}$

To find maximum height Substitute $t = \frac{7}{8}$ in the given equation

$h(t)=-16(\frac{7}{8})^2+28(\frac{7}{8})+15\\h(t)=27.25 feet$

2)

Substitute h(t)=6

So, $-16t^2+28t+15=6$

$-16t^2+28t+9=0\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\t=\frac{-28\pm \sqrt{28^2-4(-16)(9)}}{2(-16)}\\t=\frac{-28+ \sqrt{28^2-4(-16)(9)}}{2(-16)},\frac{-28- \sqrt{28^2-4(-16)(9)}}{2(-16)}\\t=-0.2774,2.027$

3)

Domain = All real numbers

Range = y∈R : $y \leq \frac{109}{4}$