# Writing Parabola Equations

• Mar 14th 2010, 01:02 PM
nuckers
Writing Parabola Equations
Please help me with this, the teachers are on spring break for another week and i would really like to figure this out. I'm thinking that there is supposed to be two x-intercepts, not an x and y.

Here is the question:

Write the equation for the parabola in the form $\displaystyle y=a(x-h)^2+k$ with a y-intercept 10, x-intercept 2, and equation of axis of symmetry $\displaystyle x-3=0$
• Mar 14th 2010, 01:19 PM
skeeter
Quote:

Originally Posted by nuckers
Please help me with this, the teachers are on spring break for another week and i would really like to figure this out. I'm thinking that there is supposed to be two x-intercepts, not an x and y.

Here is the question:

Write the equation for the parabola in the form $\displaystyle y=a(x-h)^2+k$ with a y-intercept 10, x-intercept 2, and equation of axis of symmetry $\displaystyle x-3=0$

$\displaystyle x = h$ is the axis of symmetry ...

$\displaystyle y = a(x-3)^2 + k$

$\displaystyle 10 = a(0-3)^2 + k$ ... y-intercept

$\displaystyle 0 = a(2-3)^2 + k$ ... x-intercept

solve for $\displaystyle a$ and $\displaystyle k$
• Mar 14th 2010, 01:37 PM
nuckers
Quote:

Originally Posted by skeeter
$\displaystyle x = h$ is the axis of symmetry ...

$\displaystyle y = a(x-3)^2 + k$

$\displaystyle 10 = a(0-3)^2 + k$ ... y-intercept

$\displaystyle 0 = a(2-3)^2 + k$ ... x-intercept

solve for $\displaystyle a$ and $\displaystyle k$

Thanks, but thats what i did, but when i input it into my graphing calculator i don't get the x-intercept of 2, it comes out as 2.55

I ended up with the equation $\displaystyle y=10/9(x-3)^2$
• Mar 14th 2010, 01:45 PM
skeeter
Quote:

Originally Posted by nuckers
Thanks, but thats what i did, but when i input it into my graphing calculator i don't get the x-intercept of 2, it comes out as 2.55

I ended up with the equation $\displaystyle y=10/9(x-3)^2$

$\displaystyle 10 = 9a+k$

$\displaystyle 0 = a+k$

try again.
• Mar 14th 2010, 02:36 PM
nuckers
Ha Ha
K, i think i got it, just did a few things backwards
$\displaystyle y=100/81(x-3)^2-10/9$

Thanks so much for your help(Clapping)
• Mar 14th 2010, 03:24 PM
skeeter
Quote:

Originally Posted by nuckers
Ha Ha
K, i think i got it, just did a few things backwards
$\displaystyle y=100/81(x-3)^2-10/9$

Thanks so much for your help(Clapping)

... sorry, but that is not correct.

$\displaystyle 10 = 9a + k$

$\displaystyle 0 = a+k$ ... $\displaystyle k = -a$

$\displaystyle 10 = 9a + (-a)$

$\displaystyle 10 = 8a$

$\displaystyle a = \frac{10}{8} = \frac{5}{4}$

$\displaystyle k = -a = -\frac{5}{4}$

$\displaystyle y = \frac{5}{4}(x - 3)^2 - \frac{5}{4}$
• Mar 14th 2010, 05:10 PM
nuckers
wow, i wasn't even close, not i have to figure out what i was doing wrong