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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) = x2 + 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) = 3x2 + 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) = x2 + 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) = x3. 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) = x3. 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) = x4. 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) = -2x3 + 4x2
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) = -2x3. At
x = 0 we have f(x) 4x2, 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) = x4. 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 .