SOLUTION: Find the vertex and intercepts of the following function: f(x)= 2x^2 + 4x - 5

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Question 472572: Find the vertex and intercepts of the following function:
f(x)= 2x^2 + 4x - 5
Answer by jorel1380(3719) 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 2x%5E2%2B4x%2B-5+=+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=%284%29%5E2-4%2A2%2A-5=56.

Discriminant d=56 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+56+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+56+%29%29%2F2%5C2+=+0.870828693386971
x%5B2%5D+=+%28-%284%29-sqrt%28+56+%29%29%2F2%5C2+=+-2.87082869338697

Quadratic expression 2x%5E2%2B4x%2B-5 can be factored:
2x%5E2%2B4x%2B-5+=+2%28x-0.870828693386971%29%2A%28x--2.87082869338697%29
Again, the answer is: 0.870828693386971, -2.87082869338697. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B4%2Ax%2B-5+%29
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