1. ## Help: quadratic function whose graph is given

Determining the quadratic function whose graph is given.

Graphic is a parabola going upwards.
Vertex: -3,2
Y-Intercept: 0,4

I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

f(x) = a(x-h)^2+k

2. Originally Posted by dasani
Determining the quadratic function whose graph is given.

Graphic is a parabola going upwards.
Vertex: -3,2
Y-Intercept: 0,4

I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

f(x) = a(x-h)^2+k
The coordinates of the vertex are (h,k) = (-3,2), so the quadratic equation becomes:

f(x) = a(x - (-3))^2 + 2 = a(x + 3)^2 + 2

We can use the y-intercept to find the value of "a."
We have that f(0) = 4:
4 = a(0 + 3)^2 + 2 = 9a + 2
9a = 2
a = 2/9

Therefore, the equation becomes:

f(x) = 2/9(x + 3)^2 + 2

3. Originally Posted by dasani
Determining the quadratic function whose graph is given.

Graphic is a parabola going upwards.
Vertex: -3,2
Y-Intercept: 0,4

I've tried using the following formula as the book explained, but I am not getting the right answer for some odd reason.

f(x) = a(x-h)^2+k
when a parabola is written in the form f(x) = a(x-h)^2+k, the vertex is given by (h,k) and the y-intercept (when x = 0) is given by, f(0) = a(-h)^2 + k

here we have (h,k) = (-3,2), so we have,

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

Now we also have, for the y-intercept, f(0) = 4
=> f(0) = 4 = a(3)^2 + 2
=> 4 = 9a + 2
=> a = 2/9

so our parabola is f(x) = (2/9)(x + 3)^2 + 2

EDIT: Beaten again! Thanks a lot ecMathGeek

4. That is what I got as well, however when putting into the form y=ax^2+bx+c I am gettng the wrong answer.

Can anyone assist?

5. Originally Posted by dasani
That is what I got as well, however when putting into the form y=ax^2+bx+c I am gettng the wrong answer.

Can anyone assist?
f(x) = (2/9)(x + 3)^2 + 2
=> f(x) = (2/9)(x^2 + 6x + 9) + 2
=> f(x) = (2/9)x^2 + (2/9)6x + (2/9)9 + 2
=> f(x) = (2/9)x^2 (4/3)x + 4

6. Thank you so much, that's where the problem was. I wasn't simplyfing it correctly.