SOLUTION: please help...find the root(s) of the equation using the expression x=-b/2a to find the axis of symmetry.then graph. y=x^2-x-2 thanks

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: please help...find the root(s) of the equation using the expression x=-b/2a to find the axis of symmetry.then graph. y=x^2-x-2 thanks      Log On


   

Question 141046: please help...find the root(s) of the equation using the expression x=-b/2a to find the axis of symmetry.then graph.
y=x^2-x-2 thanks
Answer by smarty1994(10) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-2+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-2=9.

Discriminant d=9 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+9+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+9+%29%29%2F2%5C1+=+2
x%5B2%5D+=+%28-%28-1%29-sqrt%28+9+%29%29%2F2%5C1+=+-1

Quadratic expression 1x%5E2%2B-1x%2B-2 can be factored:
1x%5E2%2B-1x%2B-2+=+1%28x-2%29%2A%28x--1%29
Again, the answer is: 2, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-2+%29