anonymous 5 years ago The vertex of the graph of a quadratic functon is (3,1) and the y intercept is (0,10). What is the equations of the function?

1. anonymous

$\text{Let } f(x) = k(x-a)^2 + c$ (general quadratic with square completed) (3,1) is a minimum => a = .... and c = .... When x = 0, f(x) = 10, so k = ... We can now expand and find the solution in the form f(x) = px^2 + qx + r

2. anonymous

No idea what to do after that. But thank you!

3. anonymous

Well, if a quadratic is at a minimum, x = a in the bracket (as the bracket is >= 0 for all x) So the only value is the + c This occurs at the point x = 3, y = 1, so we can deduce that the "a" = 3, and the c = 1 => f(x) = k(x-3)^2 + 1 When x = 0, f(x) = 10. So k* (-3)^2 + 1 = 10 -> 9k + 1 = 10 -> k = 1 So f(x) = (x-3)^2 + 1, which we could expand (if we want) to f(x) = ....

4. anonymous

¬_¬

5. anonymous

I didn't get it correct :/

6. anonymous

7. anonymous

(x-3)^2 + 1 = x^2 - 6x + 10, which is the answer

8. anonymous

I don't remember! :/ it's an onlne class so I can't go back and change my anwers.