Changing from quadratic formula to standard form. 
The graph of a quadratic function has $x$-intercepts $-1$ and $3$ and a range consisting of all numbers less than or equal to $4$. Determine an expression for the function.

This is my problem. I know what the graph looks like, but I only know quadratic formula and every time I input that into WolframAlpha, it combines my terms to create a different graph. I looked up the answer to this problem to check, and the answer is $-x^2+2x+3$. The only form I know is $y=af(bx+c)+d$. So how would I get that answer from this question without using $y=af(bx+c)+d$?
 A: You want a quadratic 
$$f(x)=a(x+1)(x-3)\,\,\,s.t.\,\,\,f(x)\leq 4\,\,\,\forall x\in \Bbb R\Longrightarrow$$
$$ax^2-2ax-3a-4\leq 0\,\,\,\forall x\in \Bbb R\Longrightarrow \Delta=4a^2+4a(3a+4)\leq 0\Longrightarrow $$
$$4a(a+1)\leq 0\Longrightarrow -1\leq a\leq 0$$
So choose any $\,a\,$ as above and $\,f(x)=a(x+1)(x-3)\,$ fulfills your conditions, say
$$a=-1\Longrightarrow f(x)=-x^2+2x+3$$
Another way: Since the vertex of a parabola $\,y=ax^2+bx+c\,\,,\,a\neq 0\,$ , is the point
$$\left(-\frac{b}{2a}\,,\,-\frac{\Delta}{4a}\right)\,\,,\,\Delta=b^2-4ac$$ 
and the wanted parabola obviously opens downwards (as we want a maximum point) and the vertex is its maximal point, we get with $\,f(x)=a(x+1)(x-3)=ax^2-2ax-3a\,$:
$$-\frac{\Delta}{4a}\leq 4\Longrightarrow -4a^2+12a\geq 16a\,\,(a<0\,\,\text{since the parabola opens downwards!})\Longrightarrow$$
$$4a(a+1)\leq0$$
and we get again the same solution as above.
Added: Since you want the parabola to be less than or equal to $\,4\,$ you have to take the left extreme value, i.e. $\,a=-1\,$...can you see it?
A: As DonAntonio said, You want a quadratic
$f(x)=a(x+1)(x−3)$ such that $f(x)≤4$
and $f(x) = 4$ for some $x$.
To avoid using calculus, use the fact that a parabola
has its extreme value halfway between its zeros,
so its extreme value is at 
$\frac{3+(-1)}{2} =1$.
If $a$ was positive, then $f(x)$ would get arbitrarily large
for large enough $x$. 
So $a$ must be negative.
The value of $f$ at $x = 1$ is
$a(2)(-2) = -4a$,
and we want this to equal 4,
so $a = -1$.
A: your function could be $y=a*(x+1)*(x-3)$,but because range of y is less that $4$,you can determine all values of $a$,for which it happens,consider  case  $a>0$
$x<-1$
in this case $y>0$  and $x>3$ means   $y>0$ as well,if you solve
$a*x^2-2*a*x-3*a-4<=0$;you get
$a^2+3*a^2+4*a<=0$
$4*a^2+4*a<=0$
$a*(a+1)<=0$
solution is $[-1; 0]$  because you have quadratic,$a$ must not equal to zero,we know that maximum is $4$,for this let us derivate  function ,we have
$y'=2*a*x-2*a$ 
let $y'=0$  -->$x=1$
not put $x=1$  ,we get $y(1)=-4*a$ because it is equal to $4$,it means $a=-1$
:EDITED
if you would like to make sure that  $x=1$ is maximum  and not minimum,let take second derivative test,it says that at critical point $x=x_0$ where $f'(x_0)=0$.if $f''(x_0)<0$,then $x_0$ is maximum,else minimum,in our case $f''(1)=2*a$,but because $a<0$,it means that $f''(1)<0$
