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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- Jul 7th 2008, 01:35 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 :) - Jul 7th 2008, 01:41 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.:

$\displaystyle 0 = -3x^2 + 15x + 42$

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 $\displaystyle x^{2}$:

$\displaystyle 0 = x^{2} - 5x - 14$

Now factor the quadratic to get the form: $\displaystyle 0 = (x - a)(x-b)$ and solve for x by setting $\displaystyle x - a = 0$ and $\displaystyle x - b = 0$.

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. - Jul 7th 2008, 01:43 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. - Jul 7th 2008, 01:52 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). - Jul 7th 2008, 01:55 PMbliks
- Jul 7th 2008, 01:58 PMmasters
Edit: Too slow, but here it is anyway.

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

Plot the y-intercept. In $\displaystyle y=ax^2+bx+c$, the y-intercept is "c".

Convert $\displaystyle y=ax^2+bx+c$ to vertex form and find the turnaround point. The parabola opens downward since the coefficient of $\displaystyle x^2 $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). - Jul 7th 2008, 02:02 PMbliks
- Jul 7th 2008, 02:03 PMo_O
- Jul 7th 2008, 02:08 PMbliks
- Jul 7th 2008, 02:27 PMo_O
Not quite.

$\displaystyle (x +2)(x-7) = 0$

Set each term equal to 0:

$\displaystyle x+2 = 0 \: \: \Rightarrow \: \: x = -2$

$\displaystyle x - 7 = 0 \: \: \Rightarrow \: \: x = 7$

You had your signs switched when solving for x.