Hello!
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://img.qammunity.org/2020/formulas/mathematics/middle-school/u8ncc8v133z82wmffeco2clqkny2ekfa98.png)
Where,
![a=-16\\b=64\\c=4](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5jkkqr3om4avdw9jgvmoke8umv118copuw.png)
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://img.qammunity.org/2020/formulas/mathematics/middle-school/ch6tekft92ps46aa7eudomqfwf2k53rds7.png)
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}=(-64+-√(4096+256) )/(-32)\\\\(-64+-√(4096+256) )/(-32)=(-64+-(65.96) )/(-32)\\\\t1=(-64+(65.96) )/(-32)=-0.06\\\\t2=(-64-(65.96) )/(-32)=4.0615](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jsr3vf8vahg5kb4lc9wo9yixwxw6tb77ul.png)
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://img.qammunity.org/2020/formulas/mathematics/middle-school/uqru0o158vut5rnovrcm2hz71hrm5bqbh5.png)
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!