Question 207794
You are given roots, so you can figure out the quadratic. I will do the first one, you can use the same process to do the second one
1) roots are x=-8 and x=4
You know that at the roots, the value of the function is 0
You also know that for any product to equal zero, one or more terms in that function must be 0.
Thus
x =-8
x+8 = 0
x=
4
x-4=0
Now you have the two terms for your equation
{{{(x+8)(x-4) = f(x)}}}
{{{x^2 -4x +8x -32 = f(x)}}}
{{{x^2 + 4x - 32 = f(x)}}}
This is the equation for the first problem. Since is it q quadratic, you know it is a parabola that opens up (opens up since the x^2 term is positive). You also know that the 'turning point' will be halfway between the two root. Since the roots are 12 units apart, the 'turning point' will be 6 units from each root.
-8+6 = -2
4-6 = -2
So find the value of f(x) when x=-2 and you'll have the minimum.
{{{x^2 + 4x - 32 = f(x)}}}
{{{2^2 +4*2 - 32 = f(2)}}}
{{{4 + 8 - 32 = f(2)}}}
{{{-20 = f(2)}}}
So the 'turning point is the point (2,-20)
Plot the funtion here --> http://www.wolframalpha.com/input/?i=x^2+%2B+4x+-+32+


Now do the same process for the other one. I am sure you can do it. If you still have questions, drop me an email