SOLUTION: For a parabola how to find the points on the graph of: f(x)=-3x^2+18x+5 I know I would complete the square to get f(x) = -3(x-3)+32 and from this I would have my vertex: +3,32 so

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Question 949788: For a parabola how to find the points on the graph of: f(x)=-3x^2+18x+5
I know I would complete the square to get f(x) = -3(x-3)+32 and from this I would have my vertex: +3,32 so I would shoot left 3 and up 32 and have that point I know the graph turn downwards. and to find the y-int I plug in x=0 and get (0,5) but what about the two points this parabola hits on the x-axis how do I find those points?
Thank you
Answer by macston(5194)   (Show Source): You can put this solution on YOUR website!
Set y=0 to find x-intercepts:

Solved by pluggable solver: SOLVE quadratic equation with variable
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=384 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: -0.265986323710904, 6.2659863237109. Here's your graph:

X-intercepts are at (-0.27,0) and (6.27,0).
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