SOLUTION: Quadratic relations 7) Sketch a graph of each quadratic. Label the x-intercepts and the vertex. b) y=2(x-3)(x+1) pleaseeeeeeeeeeee thank you very muchhhhhhhhhhhh

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Quadratic relations 7) Sketch a graph of each quadratic. Label the x-intercepts and the vertex. b) y=2(x-3)(x+1) pleaseeeeeeeeeeee thank you very muchhhhhhhhhhhh      Log On


   

Question 178080: Quadratic relations
7) Sketch a graph of each quadratic. Label the x-intercepts and the vertex.
b) y=2(x-3)(x+1) pleaseeeeeeeeeeee
thank you very muchhhhhhhhhhhh
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+2%28x-3%29%28x%2B1%29
The x-intercepts are the roots. The roots are values of x that make
Y=0. Looking at the equation, if x=3
y+=+2%283-3%29%28x%2B1%29
0+=+2%2A0%2A%28x%2B1%29
And if x=-1
y+=+2%28x-3%29%28-1%2B1%29
0+=+2%28x-3%29%2A0
So, x=3 and x=-1 must be the x-intercepts.
The vertex is always exactly midway between the roots, so
x%5Bv%5D+=+%28-1+%2B+3%29%2F2
x%5Bv%5D+=+2%2F2
x%5Bv%5D+=+1
So, the vertex is at (1,y%5Bv%5D)
y+=+2%28x-3%29%28x%2B1%29
y%5Bv%5D+=+2%281-3%29%281%2B1%29
y%5Bv%5D+=+2%2A%28-2%29%2A2
y%5Bv%5D+=+-8
So, the vertex must be at (1,-8)
Now I'll graph the equation to prove what I said
y+=+2%28x-3%29%28x%2B1%29
y+=+2%2A%28x%5E2+-+2x+-+3%29
y+=+2x%5E2+-+4x+-+6
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+2x%5E2+-+4x+-+6%29+