Given: <em>The height y of a ball (in feet) is given by the function </em><em>y=-1/12x^2+2x+4 </em><em>and x is the horizontal distance traveled by the ball.</em>
Part A:<em> </em><em>How high is the ball when it leaves the child's hand?</em>
Right after the ball leaves the child's hand, it has travelled 0 feet horizontally. Horizontal distance is represented by x, so we could say that x = 0.
Plug in 0 for our equation and solve for y, the height.
Part B & C: <em>How high is the ball at its maximum height?
</em>
What we basically want to do is find the vertex of the function.
There are multiple ways to do this. You could graph it or make a table, but this method is not efficient.
The method I am going to go over right now is putting the equation in vertex form.
![y=-\frac{1}{12}x^2+2x+4](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B12%7Dx%5E2%2B2x%2B4)
Move the constant to the left side.
![y-4=-\frac{1}{12}x^2+2x](https://tex.z-dn.net/?f=y-4%3D-%5Cfrac%7B1%7D%7B12%7Dx%5E2%2B2x)
Factor out the x² coefficient.
![y-4=-\frac{1}{12}(x^2-24x)](https://tex.z-dn.net/?f=y-4%3D-%5Cfrac%7B1%7D%7B12%7D%28x%5E2-24x%29)
Find out which number to add to create a perfect square trinomial.
(Half of 24 is 12, 12 squared is 144. We have to add 144/-12 (which is -12) to each side so that we end up with 144 inside the parentheses on the right side)
![y-4-12=-\frac{1}{12}(x^2-24x+144)](https://tex.z-dn.net/?f=y-4-12%3D-%5Cfrac%7B1%7D%7B12%7D%28x%5E2-24x%2B144%29)
Factor the perfect square trinomial and simplify the right side.
![y-16=-\frac{1}{12}(x-12)^2](https://tex.z-dn.net/?f=y-16%3D-%5Cfrac%7B1%7D%7B12%7D%28x-12%29%5E2)
Isolate y on the left side.
![y=-\frac{1}{12}(x+12)^2+16](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B12%7D%28x%2B12%29%5E2%2B16)
And now we are in vertex form.
Vertex form is defined as y = a(x-h)² + k with vertex (h, k).
In this case, our vertex is (12, 16).
You could've also taken the shortcut that for any quadratic f(x) = ax² + bx + c, the vertex (h, k) is (-b/2a, f(h)). That's basically a summation of this method which you can use if your teacher has taught it to you.
Part D & E: <em>What is the horizontal distance travelled by the ball when it hits the ground?</em>
When the ball hits the ground, y is going to be 0, since y is the ball's height.
There are many ways to solve a quadratic...split the middle, complete the square, and the quadratic formula.
![-\frac{1}{12}x^2+2x+4=0](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B12%7Dx%5E2%2B2x%2B4%3D0)
<u>
</u><u>Solving by splitting the midlde</u>
If your quadratic has fractions, this is not a good option.
<u>
</u><u>Solving by completing the square</u>
Move the constant over the right side.
![y=-\frac{1}{12}x^2+2x=-4](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B12%7Dx%5E2%2B2x%3D-4)
Divide by the x² coefficient.
(Dividing by -1/12 is the same as multiplying by its reciprocal, -12.)
![x^2-24x=-4\times-12](https://tex.z-dn.net/?f=x%5E2-24x%3D-4%5Ctimes-12)
Simplify the right side.
![x^2-24x=48](https://tex.z-dn.net/?f=x%5E2-24x%3D48)
Halve the x coefficient, square it, and then add it to each side.
(Half of -24 is -12, and -12 squared is 144.)
![x^2-24x+144=192](https://tex.z-dn.net/?f=x%5E2-24x%2B144%3D192)
Factor the perfect square trinomial.
![(x-12)^2=192](https://tex.z-dn.net/?f=%28x-12%29%5E2%3D192)
Take the square root of each side.
![x-12=\pm\sqrt{192}](https://tex.z-dn.net/?f=x-12%3D%5Cpm%5Csqrt%7B192%7D)
192 = 8 × 8 × 3, so we can simplify √192 to 8√3.
Add 12 to each side and we get our answer.
![x=12\pm8\sqrt{3}](https://tex.z-dn.net/?f=x%3D12%5Cpm8%5Csqrt%7B3%7D)
Our function does not apply when x or y is less than 0, of course.
12-8√3 is negative, so this cannot be our answer.
So, the ball had travelled
12+8√3 feet at the time when it hit the ground.
<u>Solving with the quadratic formula</u>
For any equation ax² + bx + c = 0, the solution for x is
![\frac{-b\pm\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
.
Our equation, y=-1/12x^2+2x+4, has a = -1/12, b=2, and c=4.
Let's plug these values into the quadratic formula.
![\frac{-2\pm\sqrt{2^2-4\cdot\frac{-1}{12}\cdot4}}{2\cdot\frac{-1}{12}}=\frac{-2\pm\sqrt{4-\frac{-4}3}}{\frac{-1}6}=\frac{-2\pm\sqrt{\frac{16}{3}}}{\frac{-1}6}=\frac{-2\pm\frac{4}{\sqrt{3}}}{\frac{-1}6}](https://tex.z-dn.net/?f=%5Cfrac%7B-2%5Cpm%5Csqrt%7B2%5E2-4%5Ccdot%5Cfrac%7B-1%7D%7B12%7D%5Ccdot4%7D%7D%7B2%5Ccdot%5Cfrac%7B-1%7D%7B12%7D%7D%3D%5Cfrac%7B-2%5Cpm%5Csqrt%7B4-%5Cfrac%7B-4%7D3%7D%7D%7B%5Cfrac%7B-1%7D6%7D%3D%5Cfrac%7B-2%5Cpm%5Csqrt%7B%5Cfrac%7B16%7D%7B3%7D%7D%7D%7B%5Cfrac%7B-1%7D6%7D%3D%5Cfrac%7B-2%5Cpm%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%7D%7D%7B%5Cfrac%7B-1%7D6%7D)
Dividing by a fraction is the same as multiplying by its reciprocal...
![-6(-2\pm\frac{4}{\sqrt{3}})=12\pm\frac{-24}{\sqrt{3}}=12\pm\frac{24}{\sqrt{3}}=12\pm\frac{24\sqrt{3}}3=\boxed{12\pm8\sqrt{3}}](https://tex.z-dn.net/?f=-6%28-2%5Cpm%5Cfrac%7B4%7D%7B%5Csqrt%7B3%7D%7D%29%3D12%5Cpm%5Cfrac%7B-24%7D%7B%5Csqrt%7B3%7D%7D%3D12%5Cpm%5Cfrac%7B24%7D%7B%5Csqrt%7B3%7D%7D%3D12%5Cpm%5Cfrac%7B24%5Csqrt%7B3%7D%7D3%3D%5Cboxed%7B12%5Cpm8%5Csqrt%7B3%7D%7D)
Of course, we only want the positive value, 12+8√3.
Revisiting Part B & C:Since parabolae are symmetrical, if you know two values of x for some value of y (like the x-intercepts we just found in part B) then you can find the average between them to find what the x value of the vertex is, then plug that in to find the y value of the vertex (the height we want)
The average between 12+8√3 and 12-8√3 is 12. Plug that in and we get 16!