<h2>Hello!</h2>
The graph is attached.
To graph a parabola we need to know the following:
- If the parabola is open upwards or downwards
- They axis intercepts (if they exist)
- The vertex position (point)
We are given the function:
![f(t)=-16t^{2}+64t+4](https://tex.z-dn.net/?f=f%28t%29%3D-16t%5E%7B2%7D%2B64t%2B4)
Where,
![a=-16\\b=64\\c=4](https://tex.z-dn.net/?f=a%3D-16%5C%5Cb%3D64%5C%5Cc%3D4)
For this case, the coefficient of the quadratic term (a) is negative, it means that the parabola opens downwards.
Finding the axis interception points:
Making the function equal to 0, we can find the x-axis (t) intercepts, but since the equation is a function of the time, we will only consider the positive values, so:
![f(t)=-16t^{2}+64t+4\\0=-16t^{2}+64t+4\\-16t^{2}+64t+4=0](https://tex.z-dn.net/?f=f%28t%29%3D-16t%5E%7B2%7D%2B64t%2B4%5C%5C0%3D-16t%5E%7B2%7D%2B64t%2B4%5C%5C-16t%5E%7B2%7D%2B64t%2B4%3D0)
Using the quadratic equation:
![\frac{-b+-\sqrt{b^{2}-4ac } }{2a}=\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}\\\\\frac{-64+-\sqrt{64^{2}-4*-16*4} }{2*-16}=\frac{-64+-\sqrt{4096+256} }{-32}\\\\\frac{-64+-\sqrt{4096+256} }{-32}=\frac{-64+-(65.96) }{-32}\\\\t1=\frac{-64+(65.96) }{-32}=-0.06\\\\t2=\frac{-64-(65.96) }{-32}=4.0615](https://tex.z-dn.net/?f=%5Cfrac%7B-b%2B-%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D%3D%5Cfrac%7B-64%2B-%5Csqrt%7B64%5E%7B2%7D-4%2A-16%2A4%7D%20%7D%7B2%2A-16%7D%5C%5C%5C%5C%5Cfrac%7B-64%2B-%5Csqrt%7B64%5E%7B2%7D-4%2A-16%2A4%7D%20%7D%7B2%2A-16%7D%3D%5Cfrac%7B-64%2B-%5Csqrt%7B4096%2B256%7D%20%7D%7B-32%7D%5C%5C%5C%5C%5Cfrac%7B-64%2B-%5Csqrt%7B4096%2B256%7D%20%7D%7B-32%7D%3D%5Cfrac%7B-64%2B-%2865.96%29%20%7D%7B-32%7D%5C%5C%5C%5Ct1%3D%5Cfrac%7B-64%2B%2865.96%29%20%7D%7B-32%7D%3D-0.06%5C%5C%5C%5Ct2%3D%5Cfrac%7B-64-%2865.96%29%20%7D%7B-32%7D%3D4.0615)
So, at t=4.0615 the height of the softball will be 0.
Since we will work only with positive values of "x", since we are working with a function of time:
Let's start from "t" equals to 0 to "t" equals to 4.0615.
So, evaluating we have:
![f(0)=-16(0)^{2}+64(0)+4=4\\\\f(1)=-16(1)^{2}+64(1)+4=52\\\\f(2)=-16(2)^{2}+64(2)+4=68\\\\f(3)=-16(3)^{2}+64(3)+4=52\\\\f(4.061)=-16(4.0615)^{2}+64(4.0615)+4=0.0034=0](https://tex.z-dn.net/?f=f%280%29%3D-16%280%29%5E%7B2%7D%2B64%280%29%2B4%3D4%5C%5C%5C%5Cf%281%29%3D-16%281%29%5E%7B2%7D%2B64%281%29%2B4%3D52%5C%5C%5C%5Cf%282%29%3D-16%282%29%5E%7B2%7D%2B64%282%29%2B4%3D68%5C%5C%5C%5Cf%283%29%3D-16%283%29%5E%7B2%7D%2B64%283%29%2B4%3D52%5C%5C%5C%5Cf%284.061%29%3D-16%284.0615%29%5E%7B2%7D%2B64%284.0615%29%2B4%3D0.0034%3D0)
Finally, we can conclude that:
- The softball reach its maximum height at t equals to 2. (68 feet)
- The softball hits the ground at t equals to 4.0615 (0 feet)
- At t equals to 0, the height of the softball is equal to 4 feet.
See the attached image for the graphic.
Have a nice day!