1. parabola intersection

Using a graphing utility to find the point of intersection of the graphs. Then confirm your solution algebraically.
y=3x^2
y=2x^2-8x-15

I did this graphically and I got (-3,27). How do I find the solution algebraically? I know it can't be hard, but I can't find it in the book.

2. Originally Posted by RenSully
Using a graphing utility to find the point of intersection of the graphs. Then confirm your solution algebraically.
y=3x^2
y=2x^2-8x-15

I did this graphically and I got (-3,27). How do I find the solution algebraically? I know it can't be hard, but I can't find it in the book.
Set $y=y$.

In otherwords, solve

$3x^2=2x^2-8x-15$

for x.

3. Can I say something though ? The question is wrong. You must give different letters to different equations. What you wrote means that for any $x$ in $R$, $3x^2=2x^2-8x-15$, which is wrong.

So I would rather denote the two parabolas like this :

$y_1 = 3x^2$
$y_2 = 2x^2 - 8x - 15$

Thus solve for $y_1 = y_2$, ... and Von Nemo already answered this I was just posting because of this (since solving $x$ for $y = y$ is absolutely meaningless, even though you understand it ... I mean let's talk math !)

4. Thanks! I wish I were studying Nietzche/philosophy instead of Pre-calc. I am never going to remember all of this. Sigh...

5. Originally Posted by Bacterius
... and Von Nemo already answered this I was just posting because of this (since solving $x$ for $y = y$ is absolutely meaningless, even though you understand it ... I mean let's talk math !)
The question - I believe - is that we are looking for $\{(x,y)\in\mathbb{R}^2|3x^2=y\text{ and }2x^2-8x-15=y\}$.

Note that this implies that $y=y$.

You are trying to distinguish between two parabolas, but notice the word "intersection". Once this word was tossed into the game, y became equal to y.

6. Oh right, thanks for the explanation