# Pre Calculus Midterm Exam 2 - Solutions

1. (10 points) Suppose that f(x) = 2x + 1 and g(x) = -x +
4.

(a) (4 points) Solve f (x) = g(x).

**Solution: **We have 2x + 1 = -x + 4 thus 3x = 3 and x = 1.

(b) (4 points) Solve f(x) < g(x)

**Solution:** We have 2x + 1 < -x + 4 thus 3x < 3 and x < 1.

(c) (2 points) What is the geometric interpretation of f(x) = g(x)?

**Solution: **f(x) and g(x) are both linear functions and their graphs are
straight lines.

The solution of f(x) = g(x) is the x-coordinate of the point of intersection of
these two

lines .

2. (20 points) Find a quadratic function whose graph is a parabola with vertex
at the

point (2,3) and whose y-intercept has the y-coordinate equal 11.

**Solution: **Since we are given the coordinates of the vertex we know that
the function

will have the form

f(x) = a(x - 2)^{2} + 3

for some a 2 R. Now, using the y-intercept we get that 11 = a(0 - 2)^{2} + 3 thus a
= 2.

3. (20 points) Consider the polynomial

P(x) = (x + 2)^{2}(x - 1)(x - 2)^{3}

(a) (3 points) List zeros of P (x) with their multiplicities.

**Solution: **We have x = -2 with multiplicity 2, x = 1 with multiplicity 1
and x = 2

with multiplicity 3.

(b) (3 points) Determine whether the graph crosses or touches the x-axis at each

x-intercept.

**Solution:** The graph touches the x-axis at x = -2 (since the multiplicity
is even) and

crosses it at x = 1 and x = 3 (since multiplicities are odd ).

(c) (10 points) Determine the behaviour of th graph near each x-intercept.

**Solution: **The polynomial P(x) is already factored. We get

At x = -2 we have

At x = 1 we have

At x = 2 we have

(d) (4 points) Determine the end behaviour, that is, find
the power function that

the graph resembles for large values of .

**Solution: **The leading term in P(x) is x^{6} thus the power function we are
looking for is

f(x) = x^{6} .

4. (30 points) Consider the functions

(a) (5 points) Find the vertical asymptotes of R(x).

**Solution: **The domain of R(x) is all real x except if x^{2} + 3x + 2 = 0.
Thus we have

two candidates for vertical asymptotes, x = -1 and x = -2. Factoring the
numerator

and denominator we have

The last formula is R (x) in lowest terms . Thus the only vertical asymptote is x
= -1.

(b) (12 points) Find the horizontal and oblique asymptotes of R(x), if any.

**Solution:** Using long division we find that

In the second term the degree of the numerator is strictly less than the degree
of the

denominator hence as we have
. y = 1 is a horizontal asymptote.

(c) (12 points) Find the horizontal and oblique asymptotes of Q(x), if any.

**Solution:** Using long division we find that

Again, in the third term the degree of the numerator is strictly less than the
degree of

the denominator hence as this term goes to
zero. y = 2x + 1 is an oblique

asymptote of Q(x)

(d) (1 point) Can a rational function have a horizontal asymptote and an oblique

asymptote at the same time? Justify your answer.

**Solution:** Every horizontal asymptote is an oblique asymptote so if a
function has a

horizontal asymptote it also has an oblique one. If a function has a
non-horizontal

oblique asymptote it cannot have another horizontal asymptote at the same time.

5. (20 points) A wire 20 meters long is to be cut into two
pieces. Both pieces will be

shaped as a square. Express the total area A enclosed by the pieces of wire as a

function of length x of a side of one of the squares. What is the domain of A?
What

is the minimal value of A? What is the maximal physically possible value of A?

**Solution:** Let the first piece have lenght 4x. The remaining part has then
length

20-4x. Since both pieces are bent to squares the first one has sides with length
x, the

second one with length 5 - x. The sum of areas of both squares is

Despite A(x) is well defined for all real x there are some restrictions i.e. 0 <
x < 5.

This is the domain of A(x). Now, notice that A(x) is a quadratic function whose
graph

is a parabola that opens up hence the minimal value is attained at the vertex.
The

x-coordinate of the vertex is h = 2.5 and this is actually the length of the
side of both

squares in this case.

Extra Credit

6. (20 points) Inscribe a right circular cylinder of height h and radius r in a
sphere of

fixed radius R (see illustration). Express the volume V of the cylinder as a
function

of h (only). (Hint: the volume of a cylinder with radius r and height h is
.
Note

also the right triangle.)

**Solution:** Let r be the radius of the base of the cylinder and h its
height. Then the

formula for the volume of a cylinder yields .
This quantity depends on two

variables -h and r. We have to get rid of r. We use the right triangle with
sides r, h/2 and

R and from the Pythagorean theorem we get .
Thus .

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