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# Precalculus I Review for Midterm

**1.** Let a(x) = 2x - 5 and b(x) = -x + 1. Solve a(x) = b(x), a(x) b(x). What are
the

graphs of a(x), b(x)? Where do the graphs of a(x) and b(x) intersect?

**Solution:** x = 2 and x≤2. Graphs of a(x), b(x) are straight lines which intersect
at

the point (2,-1).

**2.** For the function f(x) = (x + 1)^{2} - 4 find out if the graph opens up or down,
what

are to coordinates of the vertex, what are the x- and y-intercepts if any.

**Solution: **Writing out the formula we get f(x) = x^{2} + 2x - 3 which shows that
the

parabola opens up. The coordinates of the vertex are given for free - (-1,-4).
The

x-intercepts are x = 1 and x = -3. The y-intercept is (0,-3).

**3. **For the function g(x) = 3x^{2} + 4x - 1 find out if the graph opens up or down,
what

are to coordinates of the vertex, what are the x- and y-intersects if any.

**Solution:** The graph opens up. The coordinates of the vertex are (-2/3,-7/3).
The

x-intercepts are . The y-intercept is (0,-1).

**4.** Find a quadratic function that has the x- intercepts x = -2 and x = 1 and
whose

graph passes through the point (0,-2).

**Solution:** f(x) = (x + 2)(x - 1) = x^{2} + x - 2.

**5. **Find a quadratic function whose graph has the vertex at (2, 4) and has an
x-intercept

at x = 0.

**Solution:** f(x) = -(x - 2)^{2} + 4.

**6. **A landscape engineer has 200 feet of border to enclose a rectangular pond.
What

dimensions will result in the largest pond?

**Solution:** The dimensions are 50x50.

**7.** A special window in the shape of a rectangle with semicircles at each end is
to be

constructed so that the outside dimensions are 100 feet in length. Find the
dimensions

of the rectangle that maximizes its area.

**Solution: **The dimensions of the rectangle are
.

**For the following polynomials:**

a) find the x- and y-intercepts of each polynomial

b) determine whether the graph of f touches or crosses the x-axis at each x-

intercept

c) find the power function that the graph of f resembles for large values of |x|

d) determine the behaviour of f near each of the x-intercepts

e) list all zeros with their multiplicities

**8.** f(x) = x(x + 2)(x + 4)

**Solution:** The x-intercepts are x = 0, x = -2 and x = -4. The y-intercept is
(0, 0).

The graph crosses the x-axis at each intercept since the multiplicities are 1.
The power

function we're looking for is f(x) = x^{3}. At x = 0 we have f(x)8x, at x = -2 we

have f(x) -4(x + 2) and at x = -4 we have f(x)8(x + 4). The zeros are x = 0

with multiplicity 1, x = -2 with multiplicity 1 and x = -4 with multiplicity
1.

**9.** f(x) = (x - 2)(x + 2)^{2}

**Solution: **The x-intercepts are x = 2, x = -2. The y-intercept is (0,-8). The
graph

crosses the x-axis at x = 2 since the multiplicity is 1 and touches the x-axis
at x = -2

since the multiplicity is 2. The power function we're looking for is f(x) = x^{3}.
At x = 2

we have f(x)16(x - 2), at x = -2 we have f(x) -4(x + 2)^{2}. The zeros are x =
2

with multiplicity 1, x = -2 with multiplicity 2.

**10.** f(x) = (x - 1)^{2}(x + 3)(x + 1)

**Solution: **The x-intercepts are x = 1, x = -3 and x = -1. The y-intercept is
(0, 3).

The graph crosses the x-axis at x = -3 and x = -1 since the multiplicities are
1

and touches the x-axis at x = 1 since the multiplicity is 2. The power function
we're

looking for is f(x) = x^{4}. At x = 1 we have f(x)8(x - 1)^{2}, at x = -3 we have

f(x) -32(x + 3) and at x = -1 we have f(x)
8(x + 1). The zeros are x = 1 with

multiplicity 2, x = -3 with multiplicity 1 and x = -1 with multiplicity 1.

**11.** f(x) = -2x^{3} + 4x^{2}

**Solution:** The x-intercepts are x = 0, x = 2. The y-intercept is (0, 0). The
graph

crosses the x-axis at x = 2 since the multiplicity is 1 and touches the x-axis
at x = 0

since the multiplicity is 2. The power function we're looking for is f(x) =
-2x^{3}. At

x = 0 we have f(x) 4x^{2}, at x = 2 we have f(x)
-8(x - 2). The zeros are x = 0

with multiplicity 2, x = 2 with multiplicity 1.

**12.** f(x) = (x - 2)^{2}(x + 4)^{2}

**Solution:** The x-intercepts are x = 2 and x = -4. The y-intercept is (0, 64).
The

graph touches the x-axis at each x-intercept since the multiplicities are 2. The
power

function we're looking for is f(x) = x^{4}. At x = 2 we have f(x)
36(x-2)^{2}, at x =
-4

we have f(x) 36(x + 4)^{2}. The zeros are x = -4 with multiplicity 2, x = 2 with

multiplicity 2.

For the following functions nd the domain, horizontal, vertical or oblique
asymp-

totes.

**13.
**

**Solution:**Domain is . Vertical asymptotes are x = -3 and x = 3. The

horizontal asymptote is y = 0.

**14.**

**Solution:** Domain is . Vertical asymptotes is x = 2. The oblique asymptote
is

y = x + 2.

**15.**

**Solution:** Domain is . x = -2 is a vertical asymptote. y = 1 is a
horizontal

asymptote.

**16.**

**Solution: **Domain is . x = 1 is a vertical asymptote. y = 1 is a
horizontal

asymptote.

Solve the following inequalities

**17.
**

**Solution:**The solution is .

**18.**

**Solution:** The solution is .