Quadratics Question

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July 7th 2008, 01:35 PM
bliks

Quadratics 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, 01:41 PM
o_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.:
$0 = -3x^2 + 15x + 42$

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.
July 7th 2008, 01:43 PM
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.
July 7th 2008, 01:52 PM
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).
July 7th 2008, 01:55 PM
bliks

Quote:

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!

I appreciate your time. :)

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

Sorry to bother you again.
July 7th 2008, 01:58 PM
masters

Quote:

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).
July 7th 2008, 02:02 PM
bliks

Quote:

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.
July 7th 2008, 02:03 PM
o_O

Quote:

Originally Posted by bliks

Well thank you again o_O it makes sense now!

I appreciate your time. :)

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!
July 7th 2008, 02:08 PM
bliks

Quote:

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).

Quote:

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?
July 7th 2008, 02:27 PM
o_O

Not quite.

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

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

You had your signs switched when solving for x.