I was wondering if someone would be kind enough to walk me through how to do this problem:

Describe when the parabola y=-3x(power of 2) + 15x + 42 crosses the x-axis. Confirm your answer by graphing the equation.

Thank you for the help guys :)

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- July 7th 2008, 02:35 PMbliksQuadratics Question
I was wondering if someone would be kind enough to walk me through how to do this problem:

Describe when the parabola y=-3x(power of 2) + 15x + 42 crosses the x-axis. Confirm your answer by graphing the equation.

Thank you for the help guys :) - July 7th 2008, 02:41 PMo_O
Note that when a point lies on the x-axis, it has a y-coordinate of

**0**. So in order to find your roots, we set your quadratic equation equal to 0, i.e.:

First, divide -3 from both sides since we're going to have to factor and it's much easier to deal with no coefficients in front of the :

Now factor the quadratic to get the form: and solve for x by setting and .

This works because when you multiply 2 numbers to get 0, either one or the other or both are equal to 0. So, setting both terms (x-a) and (x-b) equal to 0 will lead you to your x-values that correspond to the y-intercept. - July 7th 2008, 02:43 PMbliks
Thank you!

So having it in the factored form gives me enough information to graph it as well?

Well, actually.. I don't quite understand how one would graph this out from just the factored form. - July 7th 2008, 02:52 PMo_O
Here's a smart way to think about it. You have your two x-intercepts right? This is enough information to find the x-value of your vertex since the quadratic (parabola) is symmetric about it. Your x-value for the vertex should be dead centre between your two x-intercepts.

Then, plug this x-value back into your original quadratic to find its value. Now, you have your vertex and x-intercepts which is enough to graph it. (You may also want to find the y-intercepts by simply setting x = 0 in your original quadratic). - July 7th 2008, 02:55 PMbliks
- July 7th 2008, 02:58 PMmasters
Edit: Too slow, but here it is anyway.

Plot the x-intercepts found after you factor the quadratic.

Plot the y-intercept. In , the y-intercept is "c".

Convert to vertex form and find the turnaround point. The parabola opens downward since the coefficient of is negative.

If you need more points to plot, set up a table using

x and f(x). Assign arbitrary values for x and solve for f(x). - July 7th 2008, 03:02 PMbliks
- July 7th 2008, 03:03 PMo_O
- July 7th 2008, 03:08 PMbliks
- July 7th 2008, 03:27 PMo_O
Not quite.

Set each term equal to 0:

You had your signs switched when solving for x.