SOLUTION: Hi all, I am still having trouble with some more graphing problems. I need to find the y-intercept, the x intercept(s), if any, and turning point of the function y = 2x^2

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Question 200363: Hi all, I am still having trouble with some more graphing problems. I need to find the y-intercept, the x intercept(s), if any, and turning point of the function y = 2x^2 - 8x + 9. I need to show all working and sketch the graph. Any help would be great.
I also have a seperate similiar problem where the function is y = x^2 - x - 2
Hopefully someone can give me a hand.
Thnaks, -Nick.
Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Substitute 0 for to calculate the -intercept. In both examples you provided it will be the value of the constant term in the function. In general, the -intercept for is

Set the function equal to zero and solve for . The -intercept(s) will be the real number solution(s). If the solution is a conjugate pair of complex numbers, then there are no -intercepts. This is the case with your first example. The second example has a pair of real number intercepts.

In general if , then the -intercepts of are , otherwise there are no -intercepts.

Functions of the form , where graph to a parabola. Parabolas have one turning point, namely the vertex. The -coordinate of the vertex of such a parabola is found by the calculation: .

The -coordinate is the value of the function at that -value, namely:

Here are the graphs of your two examples. First one in red, second one in green.

John

SOLUTION: Hi all, I am still having trouble with some more graphing problems. I need to find the y-intercept, the x intercept(s), if any, and turning point of the function y = 2x^2 - 8x + 9.