A quadratic function has its vertex at the point (2,8) The function passes through the point (3,1) Find the quadratic and linear coefficients and the constant term of the function.
What formula do I use? I can't remember!

Question

A quadratic function has its vertex at the point (2,8) The function passes through the point (3,1) Find the quadratic and linear coefficients and the constant term of the function. 
What formula do I use? I can't remember!

jtvatsim · Answer

Use, $$y=a(x-h)^2+k$$
where (h, k) is the vertex.

anonymous · Answer

Then how do I find the value for a?

anonymous · Answer

$$(x-h)^2=2a(y-k)$$
$$(x-2)^2=2a(y-8)$$
$$x^2-4x+4=2ay-16a$$
$$2ay=x^2-4x+4+16a$$
$$2ay=x^2-4x+4(1+4a)$$
Sub in the the point (3,1) to find a.
$$2a=3^2-4(3)+4+4a$$
$$2a=-9+12-4$$
$$2a=-1$$
$$a=-\frac{1}{2}$$
Therefore equation is
$$(x-2)^2=-(y-8)$$

anonymous · Answer

Don't know where jtvatsim got that equation from, but mine is the standard form of an equation of a parabola.

anonymous · Answer

2ay=x2−4x+4(1+4a)
In this step, where did you get (1+4a)?

anonymous · Answer

factorised the 4+16a...

anonymous · Answer

Which I didn't need to anyway.

anonymous · Answer

Oh oops. it's 4+16a.

anonymous · Answer

Let me rewrite the last few lines. Did a boo-boo.

anonymous · Answer

that's okay! One thing, though, how does one find the linear coefficients and the constant term now?

anonymous · Answer

$$2a(1)=3^2−4(3)+4+16a$$
$$2a=9-12+4+16a$$
$$14a=-1$$
$$a=-\frac{1}{14}$$
therefore
$$(x-2)^2=2(-\frac{1}{14})[y-8]$$

anonymous · Answer

Didn't you read before that I expanded/distributed the brackets in order to find a. You can use that expansion to find your constant. And the "4+16a" is your constant. You just sub in a-value to find your constant.

anonymous · Answer

$$2ay=x^2−4x+(4+16a)$$
$$y=\frac{x^2}{2a}-\frac{2x}{a}+\frac{2(1+4a)}{a}$$

anonymous · Answer

$$y=(x ^{2}/2a)-(2x/a)+(2(1+4a)/2)$$
I'm sorry for seeming like the biggest goof in the world, but I'm stuck here.

anonymous · Answer

You sub in that a-value I just found for you.

anonymous · Answer

The last term there is your constant.

anonymous · Answer

1/2a is your coefficient for x^2. And you can decide for yourself what that coefficient is for the middle term.

anonymous · Answer

What do you mean I get to decide myself?

anonymous · Answer

After giving you two answers, you should be capable of knowing the coefficient of x.

anonymous · Answer

You have been super helpful!  It's been a while since I've worked with these types of problems. I'm an English major, so this is not my forte.
Thank you!!

anonymous · Answer

Also, I actually changed the numbers in the problem by the slightest bit, so I wouldn't get the exact answers. I just needed to see the method by which the answers are achieved, not the actual answers. So don't worry, you didn't give them away!

anonymous · Answer

No worries. Anytime.