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

I tried foiling the two zeros but that doesn't give me 8 when for f(5).

Like:
(x -3)(x - 7)
x² - 10x + 21
(5)² - 10(5) + 21 = -4

?

2. Hello

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

I tried foiling the two zeros but that doesn't give me 8 when for f(5).

Like:
(x -3)(x - 7)
x² - 10x + 21
(5)² - 10(5) + 21 = -4

?
This was a good trial
You're just forgetting something : the leading coefficient.

a(x-3)(x-7) also has x-intercepts at 3 & 7 , where a is a constant (different to 0, if possible ^^)

Then use f(5)=8 to determine a.

3. 0 = a(x - 3)(x - 7)
8 = a(5 - 3)(5 - 7)
-2 = a ??

4. Originally Posted by Power
y = a(x - 3)(x - 7)
8 = a(5 - 3)(5 - 7)
-2 = a ??

Why do you think this isn't right ? It's exactly the solution

Though you can expand -2(x-3)(x-7) in order to get a quadratic form... You can choose whatever display you want^^

5. Originally Posted by Power
0 = a(x - 3)(x - 7)
8 = a(5 - 3)(5 - 7)
-2 = a ??

$\displaystyle f(x) = -2(x - 3)(x - 7)$
$\displaystyle f(3) = 0$
$\displaystyle f(7) = 0$
$\displaystyle f(5) = -2(2)(-2) = 8$