y = f(x), where the roots are 1 and 3, and the minimum point is (2,-1). sketch the graph, marking the points of intersection of the axes.

so i sketched it, but don't know how to find the y intercept, and therefore the formula :P

Question

y = f(x), where the roots are 1 and 3, and the minimum point is (2,-1). sketch the graph, marking the points of intersection of the axes. 

so i sketched it, but don't know how to find the y intercept, and therefore the formula :P

anonymous · Answer

is this a quadratic function?  degree 2 right?

anonymous · Answer

you can easily find the equation. in fact you have too much information. 
method 1: vertex is $(2,-1)$ tells you it looks like 
$$f(x)=a(x-2)^2-1$$ the only thing you don't know it $a$
but you know $f(3)=0$ so 
$$a(3-2)^2-1=0$$
$$a-1=0$$
$$a=1$$ and your function is 
$$f(x)=(x-2)^2-1$$

anonymous · Answer

method 2:
you have the zeros are 1 and 3 so it must look like 
$$f(x)=a(x-1)(x-3)$$ and again all you are missing is $a$
this time use the fact that $f(2)=-1$ so get 
$$f(2)=a(2-1)(2-3)=-1$$
$$-a=-1$$
$$a=0$$ as before, so it is 
$$f(x)=(x-1)(x-3)$$ 
expand either one to put in standard form

anonymous · Answer

if any step is not clear let me know

anonymous · Answer

well, in method 2 you meant to say -a = -1, therefore a = 1, not 0 :P but anyhoo, thanks, i understood it really well :)