Refer to this previous solution set
https://brainly.com/question/26114608
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Problem 4
Like the three earlier problems, we'll place the kicker at the origin and have her kick to the right. The two roots in this case are x = 0 and x = 20 to represent when the ball is on the ground.
This leads to the factors x and x-20 and the equation [tex]y = ax(x-20)[/tex]
We'll plug in (x,y) = (10,28) which is the vertex point. The 10 is the midpoint of 0 and 20 mentioned earlier.
Let's solve for 'a'.
[tex]y = ax(x-20)\\\\28 = a*10(10-20)\\\\28 = -100a\\\\a = -\frac{28}{100}\\\\a = -\frac{7}{25}\\\\[/tex]
This then leads us to:
[tex]y = ax(x-20)\\\\y = -\frac{7}{25}x(x-20)\\\\y = -\frac{7}{25}x*x-\frac{7}{25}x*(-20)\\\\y = -\frac{7}{25}x^2+\frac{28}{5}x\\\\[/tex]
The equation is in the form [tex]y = ax^2+bx+c[/tex] with [tex]a = -\frac{7}{25}, \ b = \frac{28}{5}, \ c = 0[/tex]
The graph is below in blue.
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Problem 5
The same set up applies as before.
This time we have the roots x = 0 and x = 100 to lead to the factors x and x-100. We have the equation [tex]y = ax(x-100)[/tex]
We'll use the vertex point (50,12) to find 'a'.
[tex]y = ax(x-100)\\\\12 = a*50(50-100)\\\\12 = -2500a\\\\a = -\frac{12}{2500}\\\\a = -\frac{3}{625}\\\\[/tex]
Then we can find the standard form
[tex]y = ax(x-100)\\\\y = -\frac{3}{625}x(x-100)\\\\y = -\frac{3}{625}x*x-\frac{3}{625}x*(-100)\\\\y = -\frac{3}{625}x^2+\frac{12}{25}x\\\\[/tex]
The graph is below in red.
