SOLUTION: Please find the Vertex, min/max, x-int, and y-int of y=-3x^2+20x-12

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Question 1069442: Please find the Vertex, min/max, x-int, and y-int of y=-3x^2+20x-12
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please find the Vertex, min/max, x-int, and y-int of y=-3x^2+20x-12
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The y-int is easy, it's -12.
It's at x = 0
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The coefficient of the x^2 term is negative --> a maximum.
The max is at the vertex.
The vertex is on the Axis of Symmetry (AOS) which is x = -b/2a
x = -20/-6 = 10/3 is the AOS
y = -3(10/3)^2 + 20(10/3 - 12 = -100/3 + 200/3 - 12 = 64/3
--> vertex and max at (10/3,64/3)
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Solve for x to find the x-ints
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -3x%5E2%2B20x%2B-12+=+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=%2820%29%5E2-4%2A-3%2A-12=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+256+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2820%29%2Bsqrt%28+256+%29%29%2F2%5C-3+=+0.666666666666667
x%5B2%5D+=+%28-%2820%29-sqrt%28+256+%29%29%2F2%5C-3+=+6

Quadratic expression -3x%5E2%2B20x%2B-12 can be factored:
-3x%5E2%2B20x%2B-12+=+%28x-0.666666666666667%29%2A%28x-6%29
Again, the answer is: 0.666666666666667, 6. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-3%2Ax%5E2%2B20%2Ax%2B-12+%29

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x = 6
x = 2/3