SOLUTION: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks. sketch the graphs,using the roots to find the function and the turning po

Algebra ->  Graphs -> SOLUTION: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks. sketch the graphs,using the roots to find the function and the turning po      Log On


   



Question 207794: i posted these problems 4 days ago,but haven't received a response.could someone please help me.Thanks.
sketch the graphs,using the roots to find the function and the turning point
1.x = -8, x = 4
2.x = -3, x = 3
thanks for the help.hagd

Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
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
%28x%2B8%29%28x-4%29+=+f%28x%29
x%5E2+-4x+%2B8x+-32+=+f%28x%29
x%5E2+%2B+4x+-+32+=+f%28x%29
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%5E2+%2B+4x+-+32+=+f%28x%29
2%5E2+%2B4%2A2+-+32+=+f%282%29
4+%2B+8+-+32+=+f%282%29
-20+=+f%282%29
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