Math 147 - Precalculus
1
September 28
Fall 2016
Exam 1 Review
1. Consider the points P (2, 3) and Q(5, −1).
(a) Find the distance from P to Q.
Solution: The distance is given by the following:
p
p
(x1 − x2 )2 + (y1 − y2 )2 = (2 − 5)2 + (3 − (−1))2
p
= (−3)2 + (4)2
√
= 9 + 16
√
= 25 = 5
(b) Identify the coordinates of the midpoint of the line segment P Q.
Solution: The midpoint has the following coordinates:
x1 + x2 y1 + y2
2 + 5 3 + (−1)
,
=
,
2
2
2
2
7
=
,1
2
√
2. Identify the domain and range of the function f (x) = 2x − 3 + 4.
103 the x-values that make 2x −
Alternate
Midterm
Solution: The domain of the Math
f includes
3 ≥ 0, i.e.
x ≥ 13/2, so the
√
domain is [3/2, ∞). The range is [4, ∞), since 2x − 3 can output any non-negative number.
1. [12 points] Consider the figure below.
3. Find an equation for the line ` in the figure below.
y
x2
f (x) = 16
-1
4
`
x
(a) Compute the average rate of change for f (x) on the interval
Solution: Using the formula for the parabola, the line ` contains the points (4, 0) and (−1, 15). The
0  x  4.
slope of ` is
y2 − y1
x2 − x1
15 − 0
=
−1 − 4
15
=
= −3
−5
m=
S
Math 147 - Precalculus
September 28
Fall 2016
Using point slope form with m = −3 and the point (4, 0), we have the following equation for the line:
y − 0 = −3(x − 4).
4. The domain of h(x) is −8 ≤ x ≤ 5 and its range is −2 ≤ h(x) ≤ 4. Determine the domain and range
of h(x + 3) − 1.
Solution: The graph of y = h(x + 3) − 1 is the graph of y = h(x) shifted down 1 unit and right 3
units. This subtracts 1 from each value in the range and subtracts 3 from each value in the domain.
The domain of h(x + 3) − 1 is −11 ≤ x ≤ 2 and the range is −3 ≤ y ≤ 3.
5. Using the graph below, evaluate the following quantities.
y
g(x)
f (x)
1
1
x
(a) g(f (−3)) Solution: Using the graph, f (−3) = 2, sog(f (−3)) = g(2) = 4.
(b) f (1)g(1) Solution: We have f (1) = −2 and g(1) = 3, so f (1)g(1) = −6.
(c) f (g −1 (1)) Solution: The value of g −1 (1) is the x-value for which g(x) = 1; the graph shows
that the graph of g contains the point (−2, 1), so g −1 (1) = −2. Then f (g −1 (1)) = f (−2) = 2.
September 28
Math 147 - Precalculus
Fall 2016
6. The function g(x) is pictured below. Graph the function g(x + 1) − 2 on the same set of axes. (Hint:
it may help to start by identifying the points where the graph changes and shifting those points.)
y
x
Solution: The graph of g(x + 1) − 2 is in red; it is the graph of g(x) shifted down 2 units and left 1
unit.
7. Match the equations (a)-(e) to the lines graphed (i)-(vii). Write the number of the graph in the blank
next to the equation to which it corresponds. Note: Not every graph goes with an equation.
You can assume that a and m are positive numbers.
(a) y = x − a
(b) y = x + a
(i)
(e) −mx − a = y
(c) x = a
(d) y = −mx + a
y
y
(ii)
x
(v)
y
(iii)
x
y
(vi)
x
y
(iv)
x
y
(vii)
x
x
y
x
Math 147 - Precalculus
September 28
Fall 2016
Solution: (a) – (v) slope 1 and negative y-intercept
(b) – (ii) slope 1 and positive y-intercept
(c) – (vii) a vertical line with positive x-value
(d) – (iv) negative slope and positive y-intercept
(e) – (vi) negative slope and negative y-intercept
8. Consider the quadratic function f (x) = −5x2 + 40x.
(a) Identify the x-intercept(s) of f .
Solution: Express f in factored form to identify the x-intercept(s):
−5x2 + 40x = − 5x(x − 8) = −5(x − 0)(x − 8)
The x-intercepts of f are 0 and 8.
(b) Identify the y-intercept of f .
Solution: The y intercept is the value of f (0); this is −5(0)2 + 40(0) = 0.
(c) Identify the vertex of f .
Solution: Express f in vertex form to identify the vertex:
−5x2 + 40x = − 5(x2 − 8x)
= − 5(x2 − 8x + 16 − 16)
= − 5 (x − 4)2 − 16
= − 5(x − 4)2 + 80
The vertex is at (4, 80).
(d) Does the position of the vertex represent the maximum or minimum value of f ? Explain your
reasoning.
Solution: Since the a value is negative (namely −4), the parabola is downward facing with its
highest point at the vertex. The position of the vertex represents the maximum value of f ; this
value is 80.