Quadratic Graph! The Hunt For The Equation.

$y=k(x-a)(x-b)$ for roots $a$ and $b$

$y=k(x-1)(x-5)$

When $x=0$ ,

$4=k(0-1)(0-5); k=\frac {4} {5}$

Thus, $y=\frac {4} {5}(x-1)(x-5)$

or

$y=ax^2+bx+c$

When $x=0$ , $4=a(0)^2+b(0)+c$ , thus $c=4$

When $x=1$ , $0=a(1)^2+b(1)+4$ ; $a+b+4=0$

When $x=5$ , $0=a(5)^2+b(5)+4$ ; $25a+5b+4=0$

Solving for $a$ and $b$ , we get $a=\frac {4} {5}$ and $b=-\frac {24} {5}$

So we come up with the equation $y=\frac {4} {5}x^2-\frac {24} {5}x+4$

or $y=\frac {4} {5}(x^2-6x+5)=\frac {4} {5}(x-1)(x-5)$

Notice that the graph includes the point (0, 4). Input x=0 to each of the choices. There is only one of them which returns a value of 4, so we are done!