Writing Parabola Equations

**nuckers** · March 14th 2010, 01:02 PM

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 $y=a(x-h)^2+k$ with a y-intercept 10, x-intercept 2, and equation of axis of symmetry $x-3=0$

**skeeter** · March 14th 2010, 01:19 PM

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 $y=a(x-h)^2+k$ with a y-intercept 10, x-intercept 2, and equation of axis of symmetry $x-3=0$

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

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

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

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

solve for $a$ and $k$

**nuckers** · March 14th 2010, 01:37 PM

Originally Posted by skeeter

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

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

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

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

solve for $a$ and $k$

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 $y=10/9(x-3)^2$

**skeeter** · March 14th 2010, 01:45 PM

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 $y=10/9(x-3)^2$

$10 = 9a+k$

$0 = a+k$

try again.

**nuckers** · March 14th 2010, 02:36 PM

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

Thanks so much for your help

**skeeter** · March 14th 2010, 03:24 PM

Originally Posted by nuckers

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

Thanks so much for your help

... sorry, but that is not correct.

$10 = 9a + k$

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

$10 = 9a + (-a)$

$10 = 8a$

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

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

$y = \frac{5}{4}(x - 3)^2 - \frac{5}{4}$

**nuckers** · March 14th 2010, 05:10 PM

wow, i wasn't even close, not i have to figure out what i was doing wrong

Thank you for your help

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