Even though the graph of h : x —> x^2 + 4x is this parabola:since it has a inverse, it must pass the horizontal line test. But as you see in the graph below, the green horizontal lines cuts the graph in two places, so the function as graphed does not have an inverse. It is not one-to-one. This is where comes in. We must chop off enough of the graph, so that what's left will pass the horizontal line test. We will chop it off at the smallest value of x so that no points on the graph will be directly to the left of any other part of the graph. So obviously we chop it off at the very bottom point, which is the vertex of the parabola. To find the vertex of a parabola we use the formula for the x-coordinate of the vertex, which is . In the equation "a" is the coefficient of x2, which is 1. The "b" is the coefficient of x, which is 4. . Therefore k=-2. So the x-coordinate of the vertex is -2, when we substitute -2 for x in we get the y-coordinate of the vertex . So the vertex is the point (-2,-4). So we chop the parabola where x=k=2, right at the vertex, so we will only leave the right half of the parabola, like this: . We only use the part at or right of x=-2 To find the inverse of h(x) = x2 + 4x for x > -2 1. We replace h(x) by y, , 2. We replace x by y and replace y by x: , We solve for y: , Multiply through by -1 to make squared term positive: , Use the quadratic formula: Now we work out the domain of the inverse Replace y by what y equals: Add 2 to both sides Since it is non-negative we use the positive sign for the square root. That tells us that for the inverse we also use the positive sign for the same square root: Going back to the inequality, we square both sides. That's the domain of the inverse function. So the inverse is although we write h-1(x) for y: <---ANSWER If you want to use the same kind of notation your teacher uses, <---ORIGINAL FUNCTION with k = -2 <---INVERSE FUNCTION Here is the graph of the inverse on the same set of axes (in blue): And you see that the inverse is the reflection of the original function across the identity line, whose equation is y = x (where x and y are identically equal and the identity line is the line that bisects the 1st and 3rd quadrants (in green, dashed since it's not part of either graph). Edwin