INFORMATION:
We know that the function that describes Katie's parabolic trajectory is
![h(t)=-16t^2+16t+672](https://img.qammunity.org/2023/formulas/mathematics/college/7a6h8diqqj9c55ab7u4xpffkre9ufjqgz2.png)
And we must graph it, identifying Katie's start and end points
STEP BY STEP EXPLANATION:
Since we have a parabola, we can calculate the vertex to graph it
The formula for the vertex is
![x=(-b)/(2a)](https://img.qammunity.org/2023/formulas/mathematics/high-school/thp2xvy4ibcymljghqfdd580j7p1ckgbwu.png)
In this case, a = -16 and b = 16
Now, replacing in the formula
![x=(-16)/(2(-16))=(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/i5a4vnn0zasl0k9gpgfwzjqst4bi1kbqq0.png)
Now, we must replace t = 1/2, to find the coordinate of the vertex
![\begin{gathered} h((1)/(2))=-16((1)/(2))^2+16\cdot(1)/(2)+672 \\ h((1)/(2))=676 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/dqvalt12jsk4ismrqwttw1fvt6b42q78zx.png)
Then, the coordinate of the vertex is (1/2, 676)
Now, we can find the x intercepts if we equal the function to 0 and solve for t
![\begin{gathered} -16t^2+16t+672=0 \\ -16(t^2-t-42)=0 \\ -16(t-7)(t+6)=0 \\ \text{ Solving for t,} \\ t=7,t=-6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/q6yoktgvmchdrikcwy5u0vwqiyv7yo5qjw.png)
Now, knowing the vertex and the x-intercepts we can graph the function
In the graph we take the positive part of the parabola because t (x-axis) represents time and the time must be positive
The end point would be when h(t) = 0
ANSWER: