Let's start with standard form for a quadratic. f(x) = ?????Zib said:The question asks me to find the equation of the quadratic function in standard form that has zeros 1+the square root of 11 and 1-the square root of 11 and that contains the point (4, -6)
Can anyone tell me how to set this up to begin with?
They are just numbers. Substitute them into a(x - u)(x - v) and do the algebra.Zib said:Thank you, but what I'm having trouble with is how 1+the square root of 11 and 1-the square root of 11 could be fit into that form. Can you help me please?
Zib said:I expand (x - u)(x - v) = ax2 - ax(u + v) + auv? Did you just miss a key
Yes -- Jeff intended to type -ax(u + v)