# Homework Help: Finding the equation of a parabola with only 2 x-intercepts and the vertex

1. Jan 29, 2012

### ashers

1. Write an equation of the quadratic function f whose graph has x-intercepts 3 and 7 and f(5) = 8.

2. I really don't know where to go from here

3. obviously, is the x-int. are 3 and 7 that means they are points (3,0) and (7,0). Also, f(5) = 8 becomes the point (5,8) which i have found to be the vertex. But, at this point, I don't know how to form a parabolic equation around those three points. Please help me!!!

2. Jan 29, 2012

### DivisionByZro

What do you know about the roots of a quadratic? You're really almost there. x=5 and x=7 are roots of f(x) and you need f(5)=8.

Hint:
Think of:

$$(x-a)(x-b)\cdot K =f(x)$$

Can you take it from here?

3. Jan 29, 2012

### ashers

i actually have no idea how to move on from there because we haven't learned that formula. I am in CP Pre-calc in high school and the simplest solution would be great. I am not very good at math and this is a review of stuff we learned from september :/

4. Jan 29, 2012

### DivisionByZro

Well if a polynomial f(x) has zeros (let's say 2 in this case, it's a quadratic), then f(x), when factored, could look like: (x-a)(x-b)*K=f(x). We say that a and b are roots of f(x) because they are zeros of the polynomial.

So you had said that (3,0) and (7,0) were solutions to f(x). Do you now see what to do?

Edit: Use the two "x-values" you have as the roots of your quadratic, and then you'll need that f(5)=8...So you will need to modify your polynomial to make it work, in other words you will need to multiply it by a factor K.

Does this help at all?

Last edited: Jan 29, 2012
5. Jan 29, 2012

### ashers

well i tried multiplying x-3 and x-7 to get the equation and i got x^2 - 10x + 21. but, if you plug in 5 as x it does't equal 8. and also if point (5,8) is the vertex that means the parabola would open downwards, which makes the equation negative. I'm still really confused though :/

6. Jan 29, 2012

### DivisionByZro

You're almost there. Just find a constant to multiply f(x)=x^2 - 10x + 21 so that f(5)=8.

7. Jan 29, 2012

### ashers

thanks!!

8. Jan 29, 2012

### ashers

so wait now all i have to do is multiply the entire formula by one constant until f(5)=8?

9. Jan 29, 2012

### DivisionByZro

Well whatever polynomial f(x) has f(5)=8 must pass through (5,8).

10. Jan 29, 2012

### ashers

but i still can't seem to figure out how to make X^2 - 10x + 21 pass through (5,8)

11. Jan 29, 2012

### DivisionByZro

With f(x)=x^2 - 10x + 21, what is f(5)?

12. Jan 29, 2012

### ashers

13. Jan 29, 2012

### ashers

so should i multiply it out by -2 so that the answer becomes 8?

14. Jan 29, 2012

### eumyang

Just plug in 5 into
f(x) = k(x2 - 10x + 21):
8 = k(5^2 - 10(5) + 21)
... and solve for k.

EDIT: Oops, a little too late. ;)

15. Jan 29, 2012

### DivisionByZro

You got it!

16. Jan 29, 2012

### ashers

THANK YOU SO MUCH! I realized i was making a multiplication error early on in the problem and this is why i could not solve it correctly. You're explanation helped so much! thanks again!