What is the quadratic function that is created with roots at 2 and 4 and a vertex at (3, 1)?

Question

anonymous · Answer

$$f(x)=a(x-2)(x-4)$$ and solve for a via 
$$f(3)=1=a(1-2)(1-3)$$

anonymous · Answer

gives 
$$1=2a$$ or 
$$a=\frac{1}{2}$$

anonymous · Answer

you could have also written 
$$f(x)=a(x-3)^2+1$$ and solved for "a" by writing 
$$f(2)=0=a(2-3)^2+1$$
hold the phone, i get a different "a"

anonymous · Answer

are you sure this question is written correctly??

anonymous · Answer

let me see

anonymous · Answer

it is...

anonymous · Answer

maybe my algebra is messed up, let me try again

wasiqss · Answer

satellit plz try my question again

anonymous · Answer

go away! he's mine!

wasiqss · Answer

anonymous · Answer

Thieves!

anonymous · Answer

oh yeah my algebra was messed up sorry

anonymous · Answer

$$f(3)=1=a(3-2)(3-4)=1$$ so 
$$-a=1$$ ir 
$$a=-1$$ just like with 
$$f(2)=0=a(2-3)^2+1$$
$$0=a+1$$
$$a=-1$$ sorry

anonymous · Answer

so answer is 
$$f(x)=-(x-2)(x-4)=-(x-3)^2+1=-x^2+6x-8$$

anonymous · Answer

Awesome! Thank you for your help and explaining everything! Long time no see man!