1. ## Parabola

Please someone help. I have summer school and I got a Take-Home Quiz. My teacher knows about the whole, google or yahoo the answer thing, so he does not care. Anyways one of the questions is this: Write an equation of a parabola with vertex (-2,3) and passing through (0,15). I have tried to get this equation, but failed. Please help.

2. You could try $y=a(x-h)^{2}+k$

(h,k) is the vertex and (x,y) are the coordinates it passes through.

Use these to find a and you're all set.

3. How would I use it? Would i enter the vertex and the other thing into it?

Sorry ... me = slow

4. Originally Posted by iorga
Please someone help. I have summer school and I got a Take-Home Quiz. My teacher knows about the whole, google or yahoo the answer thing, so he does not care. Anyways one of the questions is this: Write an equation of a parabola with vertex (-2,3) and passing through (0,15). I have tried to get this equation, but failed. Please help.
$y = a(x - h)^2 + k$

you want, $y = 15$, $x = 0$, $h = -2$ and $k = 3$

plug these into the formula to find $a$

Once you've found $a$, rewrite the original equation, this time, filling in all the values except $x$ and $y$ and you will have your parabola

5. Given a parabola $y=ax^2+bx+c, a\neq 0$, its vertex has the coordinates $\left(-\frac{b}{2a},-\frac{\triangle}{4a}\right)$, where $\triangle =b^2-4ac$.
Now, we have
$\displaystyle \left\{\begin{array}{ll}\displaystyle -\frac{b}{2a}=-2\\\displaystyle -\frac{\triangle}{4a}=3\end{array}\right.$
Because the parabola is passing through the point $(0,15)$, replacing $x=0,y=15$ in the equation, we have $c=15$.
Now, the system becomes
$\left\{\begin{array}{ll}b=4a\\b^2-48a=0\end{array}\right.\Rightarrow a=3,b=12$.
So the parabola is $y=3x^2+12x+15$