SOLUTION: Hi all, I am having a meltdown with Quadratic equations, vertexes and intercepts. Could someone please help me uunderstand this problem? Find the vertex and intercepts for th

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Question 18332: Hi all,
I am having a meltdown with Quadratic equations, vertexes and intercepts. Could someone please help me uunderstand this problem?
Find the vertex and intercepts for the quadratic function , and sketch its graph:
y = x^2 - 4x -12
Thank you very much!
Answer by Earlsdon(6287) About Me  (Show Source):
You can put this solution on YOUR website!
Once your quadratic equation is in the standard form:ax%5E2+%2B+bx+%2B+c%5D=+0, the x-coordinate of the vertex is given by -b%2F2a.
-b%2F2a+=+-%28-4%29%2F2%281%29 = 4%2F2+=+2
Now find the y-coordinate of the vertex by substituting this value of x (x=2) into the quadratic equation and solve for y.
y+=+%282%29%5E2+-+4%282%29+-+2
y+=+4+-+8+-+12
y+=+-16
The vertex is at (2, -16)
The x-intercepts can be found by solving the quadratic equation.
x%5E2+-+4x+-+12+=+0 Solve by factoring.
%28x%2B2%29%28x-6%29+=+0 Apply the zero products principle.
x%2B2+=+0 and/or x-6+=+0
If x+2 = 0, then x = -2
If x-6 = 0, then x = 6
The x-intercepts are: (-2, 0) and (6, 0)
See the graph below:
graph%28300%2C200%2C-10%2C10%2C-20%2C10%2Cx%5E2-4x-12%29