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 $y = ax(x-20)$

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'.

$y = ax(x-20)\\\\28 = a*10(10-20)\\\\28 = -100a\\\\a = -\frac{28}{100}\\\\a = -\frac{7}{25}\\\\$

This then leads us to:

$y = ax(x-20)\\\\y = -\frac{7}{25}x(x-20)\\\\y = -\frac{7}{25}x*x-\frac{7}{25}x*(-20)\\\\y = -\frac{7}{25}x^2+\frac{28}{5}x\\\\$

The equation is in the form $y = ax^2+bx+c$ with $a = -\frac{7}{25}, \ b = \frac{28}{5}, \ c = 0$

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 $y = ax(x-100)$

We'll use the vertex point (50,12) to find 'a'.

$y = ax(x-100)\\\\12 = a*50(50-100)\\\\12 = -2500a\\\\a = -\frac{12}{2500}\\\\a = -\frac{3}{625}\\\\$

Then we can find the standard form

$y = ax(x-100)\\\\y = -\frac{3}{625}x(x-100)\\\\y = -\frac{3}{625}x*x-\frac{3}{625}x*(-100)\\\\y = -\frac{3}{625}x^2+\frac{12}{25}x\\\\$

The graph is below in red.

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3 years ago