Quandaries and Queries


12 grade, algebra II

Given the vertex (4, -2)
y intercept = -6
find if/where the parabola crosses the x axis?

I'm subbing for a math class, and just can't remember.

Thanks for any help you can offer.



Hi Delores,

First plot the two points (4,-2) and (0,-6) on the plane and then draw a rough sketch of what the parabola.

This is a parabola that opens downward. I know that the "standard" parabola with vertex at (0,0) and which opens downward has equation

y = - x2

The parabola you want is shifted 4 units to the right so the "x" factor should be (x - 4). The graph is then shifted up or down (in this case down) and then scaled (either expanded or contracted) to give the parabola you want. Hence I know that the equation is

y = - a(x - 4)2 + b

Now you can determine a and b since the parabola passes through (4,-2) and (0,-6).

-2 = - a(4 - 4)2 + b = b

so b = -2, and

-6 = - a(-4)2 - 2

so a = 1/4.

I don't know what detail you require to actually answer the question. It may be that the sketch at the beginning is enough to see that the parabola does not cross the x-axis.




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