Question 1069442
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
*[invoke solve_quadratic_equation -3,20,-12]
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x = 6
x = 2/3