Answer:
• (a)See graph below
,
• (b)The rock will be 258 feet high at t=2.39 seconds and at t=3.61 seconds.
Explanation:
Given the equation modeling the height, h(t) of the rock after t seconds:
[tex]h(t)=-16t^2+96t+120[/tex]
Part A
To sketch the graph, we use the intercepts and the vertex.
When t=0:
[tex]\begin{gathered} h(0)=-16(0)^2+96(0)+120 \\ h(0)=120\text{ feet} \\ \implies(0,120) \end{gathered}[/tex]
The y-intercept is (0,120)
When h(t)=0:
[tex]\begin{gathered} h(t)=-16t^2+96t+120=0 \\ \text{ We solve for t using the quadratic formula} \\ t=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\ t=\dfrac{-96\pm\sqrt{96^2-4(-16)(120)}}{2(-16)}=\dfrac{-96\pm\sqrt{16896}}{-32} \\ t=\dfrac{-96+\sqrt{16896}}{-32}\text{ or }t=\dfrac{-96-\sqrt{16896}}{-32} \\ t=-1.06\text{ or }t=7.06 \end{gathered}[/tex]
The only possible x-intercept is (7.06, 0).
To find the vertex, we use the vertex formula below:
[tex]\begin{gathered} (h,k)=\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right) \\ =\left(-\frac{96}{2(-16)},\frac{4(-16)(120)-96^2}{4(-16)}\right) \\ =(3,264) \end{gathered}[/tex]
Use these points to sketch the graph as shown below:
Part B
When the rock is 258 feet high: h(t)=258
[tex]\begin{gathered} -16t^2+96t+120=258 \\ -16t^2+96t+120-258=0 \\ \implies-16t^2+96t-138=0 \end{gathered}[/tex]
We solve the equation for t:
[tex]\begin{gathered} \text{ We solve for t using the quadratic formula} \\ t=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a} \\ a=-16,b=96,c=-138 \\ t=\dfrac{-96\pm\sqrt{96^2-4(-16)(-138)}}{2(-16)}=\dfrac{-96\pm\sqrt{384}}{-32} \\ t=\dfrac{-96+\sqrt{384}}{-32}\text{ or }t=\dfrac{-96-\sqrt{384}}{-32} \\ t=2.39\text{ or }t=3.61 \end{gathered}[/tex]
The rock will be 258 feet high at t=2.39 seconds and at t=3.61 seconds.
