How to determin the equation of quadratic function?

1. Jul 25, 2009

cruisx

Hi guys whats up. I just have a small question i wish to ask. If you are given vertex(-2,3) and it passes through point(2,5) how do you determine its equation? Its been around 6 months since i last did some math so i am having a hard time remembering. I am not asking you guys to solve it but just to show me how to tackle it =). thanks.

2. Jul 25, 2009

mathman

I presume you are asking for a curve of the form y=ax2+bx+c.
The vertex at (-2,3) gives you an equation y=a(x+2)2+3.
Plug in the point value gives you 5=16a+3 or a=1/8.

3. Jul 26, 2009

cruisx

hi, thanks for the reply but when a friend helped me out this is how we did it.

we used f(x) = a(x+2) + 3

then we sub'ed in our point (2,5) to get

5 = a(2+2)^2 + 3
5 = a(4)^2 + 3
5 = 16a + 3
2 = 16a
a = 8

after that we just put a back in to the equation to get

f(x) = 8(x+2)^2 + 3

is this correct?

Also another question i wanted to ask is how would i get the quadratic equation if you are given an X coordinate of the vertex in 2, a = -1 and passes through point (3,4)? This is what i am thinking but i dont know if this is correct.

f(x) = -1(x -2 )2 + 0
4 = -1(3-2)2 + 0

and then i woudl go on from there?

Last edited: Jul 26, 2009
4. Jul 26, 2009

rasmhop

Yes.

$$(1/8)x^2 + (1/2)x + 7/2 = 1/8 (x+2)^2+3$$
so it's the same answer mathman gave you.

You can check it in several ways (this is always a good exercise if you aren't completely sure of your answer). We need to check that (-2,3) and (2,5) satisfy your equation so we insert x=-2 and x=2 to check the values of y:
$$1/8*(-2)^2 + 1/2 * (-2) + 7/2 = 3$$
$$1/8*(2)^2 + 1/2 * 2 + 7/2 = 5$$
Thus you graph passes through both points. You also need to verify that it has a minimum or maximum at x = -2. This can be checked in a number of ways. In general for an equation of form:
$$ax^2 + bx + c = y$$
the minimum or maximum is at x = -2a/b. If you don't know this formula, but know a little calculus you can determine local maximum and minimum points by differentiating and setting the derivative equal to 0. This gives us:
$$1/4x + 1/2 = 0$$
in your case which has the solution x = -2. Since we know that a parabola has exactly one minimum or maximum point and it's differentiable at all points we know that this is it so the vertex is at x = -2. Otherwise you could try to complete the square to get:
$$y = 1/8(x+2)^2 + 3$$
which shows us that $$y\geq 3$$ because the square of a real number is non-negative. Hence the vertex is at y= 3 which verifies our answer of (-2,3) for the vertex.

EDIT: Apparently you edited your post while I replied and changed the correct answer to an incorrect answer. In the first formula you omitted ^2, but I expect this is a typo as you remember it in later formulas. After 2 = 16a you should get 1 = 8a which gives a= 1/8 as mathman pointed out, but you somehow got a=8 which is incorrect (16*8 is not 2). Apart from that you should read mathman's reply as his solution is exactly the same.

5. Jul 26, 2009

cruisx

ya that ^2 was a mistake, i forgot to put it there. Thanks for the help though, your explanation makes things more clearer. Its been a while so i kinda forgot how to do this but your explanation helped me out =).

Also since i changed it to the wrong equation i am guessing i am doing this one wrong as well?

you are given an X coordinate of the vertex in 2, a = -1 and passes through point (3,4)

would the equation be this: y = - x² + 9x - 14?

thanks, help would be appreciated.

Last edited: Jul 26, 2009
6. Jul 26, 2009

rasmhop

You did everything right except for the fact that you at some point got the equation 2= 16a and concluded that a =8. This step is not correct. You should have solved for a and gotten a = 1/8 and inserted that and then you would have gotten the correct answer.

No. The vertex is at the x-coordinate -9/(2(-1)) = 9/2 which is not 2. In general the vertex has an x-coordinate of -b/2a where b and a are the coefficients of x and x^2.

You are told that the x-coordinate of the vertex is 2 so you know it's of the form:
f(x) = a(x-2)^2 + c
where c is the y-coordinate of the vertex. You are told that a=-1 (I'm assuming that this is what was meant by a) so inserting this we get:
f(x) = -(x-2)^2 + c
Now inserting the point we get:
4 = -(3-2)^2 + c
4 = -(1)^2 + c
4 = -1 + c
c = 5
Thus our function is:
f(x) = -(x-2)^2 + 5
Our in expanded form:
f(x) = -x^2 + 4x + 1

7. Jul 26, 2009

cruisx

Ok thanks =) i will give it a go and see if i get the same answer again. Also just a graphing Question...maybe i am just over analyzing things but.. I have to graph/create a family of functions with the same X coordinate and with the initial function of y = X squared.
so i could graph some thing like this. (0,1) (0,2) (0,3) ??? is it that easy? I have never heard of family of functions untill today so idk.