SOLUTION: I need your help. Can someone please graph this for me? Find the vertex and intercepts for each parabola. Sketch the graph. g(x) = x^2 + x - 6 By completing the square:

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need your help. Can someone please graph this for me? Find the vertex and intercepts for each parabola. Sketch the graph. g(x) = x^2 + x - 6 By completing the square:       Log On


   

Question 37687: I need your help. Can someone please graph this for me?
Find the vertex and intercepts for each parabola. Sketch the graph.
g(x) = x^2 + x - 6

By completing the square:



Vertex = (-0.5,-6.25). y-intercept = -6
Answer by Alwayscheerful(414) 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%2B1x%2B-6+=+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=%281%29%5E2-4%2A1%2A-6=25.

Discriminant d=25 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+25+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+25+%29%29%2F2%5C1+=+2
x%5B2%5D+=+%28-%281%29-sqrt%28+25+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B1x%2B-6 can be factored:
1x%5E2%2B1x%2B-6+=+1%28x-2%29%2A%28x--3%29
Again, the answer is: 2, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-6+%29



The formula to find the vertex is -b%2F2a. That formula will give you the x value of the vertex. To find the y value, plug x back into the equation.

-b%2F2a=x
-1%2F2%281%29=x
-1%2F2=x
Plug back into the original equation
y=1%28-1%2F2%29%5E2%2B1%28-1%2F2%29-6
y=1%281%2F4%29%2B%28-1%2F2%29-6
y=.25-.5-6
y=-.25-6
y=-6.25

To find the y-intercept, you must set x equal to 0
y=1%280%29%5E2%2B1%280%29-6
y=-6
Hope this helps!