SOLUTION: The function h(x) is defined by h(x) = x^2 - 6x + 20 {x>or=a}. Given that h(x) is a one-to-one function find the smallest possible value of the constant a.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The function h(x) is defined by h(x) = x^2 - 6x + 20 {x>or=a}. Given that h(x) is a one-to-one function find the smallest possible value of the constant a.      Log On


   

Question 211084: The function h(x) is defined by h(x) = x^2 - 6x + 20 {x>or=a}.
Given that h(x) is a one-to-one function find the smallest possible value of the constant a.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
All functions have only one y-value for each x-value (or else they would not be functions). A one-to-one function also has only one x-value for each y-value. In other words a one-to-one function passes not only the vertical line test, it also passes a horizontal line test, i.e. the function intersects horizontal lines at only one (or zero) points.

Since the function you have is a parabola,
graph%28300%2C+300%2C+-4%2C+16%2C+-2%2C+18%2C+x%5E2+-6x+%2B+20%29

the only way for it to pass a horizontal line test is if the domain limits the function to all or part of the left or right "half" of the parabola. Since the problem states that x+%3E=+a, then we want all or part of the right half. And since we want the smallest possible "a", we want the entire right half of the parabola. For the entire right side we need to know the x-coordinate of the vertex.

In general the x-coordinate of the vertex of f%28x%29+=+ax%5E2+%2B+bx+%2B+c is -%28b%2F%282a%29%29. In this problem, a = 1 and b = -6. So the x-coordinate of the vertex is -%28-6%29%2F%282%281%29%29+=+6%2F2+=+3. So if a+%3E=+3, h(x) is part or all of the right half of the parabola, making h(x) a one-to-one function. So the smallest possible "a" which makes h(x) a one-to-one function is 3.