Question 166605
You are absolutely correct in what you have said!
If x = 5 and x = 1 are zeros of the quadratic equation then (x-5) and (x-1) are factors, so their product will give you the equation:
{{{(x-5)(x-1) = x^2-6x-5}}}
So that you would have:
{{{f(x) = x^2-6x+5}}} and, of course:
{{{f(0) = 0-0+5}}}
{{{f(0) = 5}}} Not 1!
But, I'm wondering if something didn't get turned around here and what was really intended was:
{{{f(1) = 0}}} ...and this is certainly true!