SOLUTION: 1.) f(x)= x2 - 1 f of X= x sqared minus one Graph each function.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 1.) f(x)= x2 - 1 f of X= x sqared minus one Graph each function.       Log On


   

Question 39478: 1.)
f(x)= x2 - 1
f of X= x sqared minus one
Graph each function.
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+x%5E2+-+1
Should look like the parabola below:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B0x%2B-1+=+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=%280%29%5E2-4%2A1%2A-1=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+1
x%5B2%5D+=+%28-%280%29-sqrt%28+4+%29%29%2F2%5C1+=+-1

Quadratic expression 1x%5E2%2B0x%2B-1 can be factored:
1x%5E2%2B0x%2B-1+=+1%28x-1%29%2A%28x--1%29
Again, the answer is: 1, -1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B0%2Ax%2B-1+%29