5. A student throws a math book from a cliff 200 feet
above water at an
inclination of 45 degrees to the horizontal, with a velocity of 100 feet
per second. The height h of the book is given by
where x is the horizontal distance of the book from the base of the cliff.
(a) How far from the base of the cliff is the height of the book at a
maximum?
The problem tells us that h(x) is the height of the book, and x
re presents its distance away from the cliff. Notice that h(x) is
a quadratic function with 
, b = 1, and c =
200.
Therefore, if we want to find the point at which the height is a
maximum, we are looking for the x coordinate of the vertex of
the graph of h. This x coordinate can be found using the formula
. So, the horizontal distance that we are
looking for is
feet.
(b) Find the maximum height of the book.
Here, we need to find the maximum value of the quadratic func-
tion, h(x). This means that we need to find the y coordinate of the
vertex. Using our value in part (a), we find the maximum height
by finding h(156.25) ≈ 278.125, so the book reaches a maximum
height of about 278.125 feet.
(c) How far from the base of the cliff will the book strike the water?
The book will strike water when h(x) = 0. We use the quadratic
formula,
, to find the x for which this
occurs.
Substituting the values of a, b, and c given in the solution to part
(a), we find that h(x) = 0 when x ≈ −138.56 or x ≈ 451.06.
Since the distance away from the cliff can’ t be negative , we omit
the first solution and conclude that h(x) = 0 when x ≈ 451.06.
This means that when the book is about 451.06 feet away from
the cliff, the book will strike the water.
6. Let f(x) = x2 + 4x + 3.
(a) Find the average rate of change of f from x = 1 to x = 2. (Re-
call that the average rate of change of f from x to c is given by
, where x ≠ c).
The formula for average rate of change is given in the problem.
To find the average rate of change of f(x) from x = 1 to x = 2,
we compute 
. Substituting x = 1 into the
equation,
we have that f(1) = 8. Also, f(2) = 15. Therefore,
(b) Using your answer in part (a), find the equation of the line that
is secant to the graph of f (x) and passes through the point (1, 8).
The slope of the secant line between x = 1 and x = 2 IS the
average rate of change from x = 1 to x = 2. In order to find
the equation of a line, we need its slope and a point on the line.
We found the slope in part (a), so all we need is a point on the
line, which is given. Using point slope form, 
,
we have that y − 8 = 7(x − 1). Simplifying, we find that the
equation for the secant line that passes through the point (1, 8) is
y = 7x + 1.
(c) Simplify the difference quotient 
The expression f(x+h) tells us to plug in x+h for x in the
equation that defines f. This does NOT tell us to take
f(x) and add h to it .
First, plug in x + h for x.
This gives us f(x + h) = (x + h)2 + 4(x + h) + 3.
We do NOT say that (x + h)2 = x2 + h2.
Instead, multiply (x + h)(x + h) using FOIL to find that
(x + h)2 = x2 + 2xh + h2.
Now, distribute the 4 to obtain that 4(x + h) = 4x + 4h.
Putting all of this together, we have that
f(x + h) = x2 + 2xh + h2 + 4x + 4h + 3. Again, this is NOT the
same as adding h to f(x), which would give us x2 + 4x + 3 + h.
Now, we have to find f(x + h) − f(x). Using what we just found
out about f(x + h), we see that
From here, we have all of the information that we need to
simplify
the difference quotient .
So, the final answer is 2x + h + 4.
7. Find the function that is finally graphed after the
following transfor-
mations are applied to the graph of 
. In each
step , write the
resulting equation and sketch its graph.
(a) Shift up 2 units.
The square root function looks like this:
(Note: This is the general shape, graphed for only a few
x-values.)
To shift the graph up 2 units, we add 2 to every y value. So, the
new equation would be 
.
The new graph has the same general shape but has been shifted
up 2 units. The following is a general idea of what your graph
should look like:
(b) Reflect about the x-axis.
Now, we want to reflect the graph found in part (a) about the x-
axis. When reflecting about the x-axis, all y values are multiplied
by −1. Therefore, the equation in this situation is
. The corresponding graph should look like
the
following:
(c) Reflect about the y-axis.
When reflecting about the y-axis, x values change sign . So, in this
case, we would plug in −x for each x value in the equation in part
(b). Therefore, the final equation after all 3 transformations are
performed is 
. A reflection of the graph in
(b)
over the y-axis will generally look like the following graph:
