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

2. 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.:
$0 = -3x^2 + 15x + 42$

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 $x^{2}$:
$0 = x^{2} - 5x - 14$

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

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.

3. 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.

4. 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).

5. Originally Posted by o_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).
Well thank you again o_O it makes sense now!

How do you figure out how high along the y-axis the vertex is though?

Sorry to bother you again.

6. Originally Posted by bliks
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.
Edit: Too slow, but here it is anyway.

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

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

Convert $y=ax^2+bx+c$ to vertex form and find the turnaround point. The parabola opens downward since the coefficient of $x^2$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).

7. Originally Posted by masters
Edit: Too slow, but here it is anyway.

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

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

Convert $y=ax^2+bx+c$ to vertex form and find the turnaround point. The parabola opens downward since the coefficient of $x^2$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).
That makes sense thank you so much for your time.

8. Originally Posted by bliks
Well thank you again o_O it makes sense now!

How do you figure out how high along the y-axis the vertex is though?

Sorry to bother you again.
Don't be sorry

Remember, anything along the y-axis has a x-coordinate = 0. So, set x = 0 in your original quadratic equation and solve for y.

Edit: Ah too slow too!

9. Originally Posted by masters
Edit: Too slow, but here it is anyway.

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

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

Convert $y=ax^2+bx+c$ to vertex form and find the turnaround point. The parabola opens downward since the coefficient of $x^2$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).
Originally Posted by o_O
Don't be sorry

Remember, anything along the y-axis has a x-coordinate = 0. So, set x = 0 in your original quadratic equation and solve for y.

Edit: Ah too slow too!

Again thank you!

So just to be sure... Y= 42 right? and the factored form comes out to be (x+2)(x-7), meaning that the zeros are -7 and 2 and the axis of symmetry would be -1?

10. Not quite.

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

Set each term equal to 0:
$x+2 = 0 \: \: \Rightarrow \: \: x = -2$
$x - 7 = 0 \: \: \Rightarrow \: \: x = 7$