Part (a)
You can google complete the square if you want to be reminded of how to convert this given quadratic equation into the parabola form they request, but I don't think that you will ever have to do something like
this again. The h and k they are referring to are the x and y coordinates, respectively, of the vertex of the parabola. In general, for any quadratic equation (ax^2 + bx + c), the x coordinate of the vertex will be -b divided by (2a). Which is to say, h = - b / (2a). Once you know h, you can sub it into the given quadratic equation to compute k, the
y-coordinate of the vertex.
Once you know h and k, sub them into the given format and you have answered part (a).
I recommend that you sketch a graph of this parabola and include it in your solution to help visualize the problem. Just make a quick table of values. Plot the vertex you just found, and choose one or two other points on each side of the
vertex.
Part (b)
The vertical line test checks if a graph is a function or not. If a vertical line can never pass through more than one point on the graph, the graph represents a function
The horizontal line test checks if a graph's inverse would be a function. If a horizontal line can never pass through more than one point on the graph, then its
inverse would be a function.
A graph is a one-to-one function if and only if it passes both the vertical and horizontal line tests. That proves that for any one value of x there is only one value of y on the graph. The function is one-to-one.
Sketch the graph of the parabola, and it is obvious that it fails the horizontal line test. So you must restrict the domain to
make it one-to-one. The most logical thing to do is to restrict the domain to all the x values starting from the vertex and to the right.
Part (c)
Be sure to use the (h, k) form they had you generate in part (a). Much easier to do the algebra with that one.
I show you how to find the inverse of a function in Lesson 8 of my book (just after
question 4 in the lecture, page 230). I like to immediately have x and y change places, then proceed to isolate y.
Once you have isolated y, make sure you replace y with f^-1 (x), the f-inverse symbol.
Since an inverse changes the x and y around, it is sort of rotating a graph sideways (and flipping it). What was vertical becomes horizontal, and vice-versa. Thus, a graph that passes
the horizontal line test, will have an inverse that passes the vertical line test, proving that the inverse will be a function. Only one-to-one functions will have inverses that are also function.
Part (d)
The domain and range for f(x) should be obvious from the sketch you made earlier in the question (but, remember, you must use the restricted domain you identified in part (b) for the one-to-one function).
Part (e)
The domain and range of the inverse function are easy. The domain of f is the range of f-inverse, and the range of f is the domain of f-inverse. Just reverse the answers for part (d).