all polynomials have domain of all reals
No polynomial has any kind of asymptote
x^2-6x+5 = (x-3)^2 - 4
So you can find the vertex, which is a minimum.
Since the vertex is farther away from x=-1 than it is from x=6, f(-1) > f(6)
Since f(3-x) = f(3+x), it is evidently not 1:1.
find the domain, range, y and x-intercept, vertical asymptote, maximum, minimum, one-to-one, and horizontal asymptote of:
a. f(x)=x^2-6x+5
b. f(x)=x^2-6x+5 on domain [-1,6]
1 answer
1 answer