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

Question 1069442: Please find the Vertex, min/max, x-int, and y-int of y=-3x^2+20x-12
Answer by Alan3354(69443) (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
===========================
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 (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=256 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.666666666666667, 6. Here's your graph:

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x = 6
x = 2/3
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