Download AP Calculus AB Summerwork

AP Calculus AB Summer work
An AP program has been established in the math department in order to assist those students
who have been successful in math in achieving the highest level that can be attained. Attached you will
find the summer work for your AP Calculus AB class. This is a review of Algebra 2 / Precalculus and
some Trigonometry concepts that you need to be proficient at before starting the more rigorous course
of AP Calculus AB. You are responsible to try all of the problems and show all work. Every problem
must be attempted even if the final answer was not obtained. Please use the following websites for
tutorials, should you have difficulty:
www.khanacademy.org/
www.patrickjmt.com/
These mandatory assignments are due the first day of school (please email with any concerns
about this). The summer work concepts will be tested the first week of school. Anyone who scores
below a 70% on the test will be placed into a regular class. The summer work is worth 1000points of
your first semester grade.
We will check my email periodically during the summer, so please don’t hesitate to contact us at:
[email protected] or
[email protected]
Thank you and have a great summer,
AP Calculus Team
Calculus AB Summerwork
Equations of Lines → Why do it? Writing equations of tangent lines to curves is a fundamental idea → students must be
comfortable with different versions of lines and manipulating equations.
1. Write in slope-intercept form:
a) Y – k = m(X – l)
b) X/a + Y/b = 1
c) aX + bY = c
Equations of Lines → Why do it? Writing equations of lines parallel and perpendicular to curves is a fundamental idea →
students must be comfortable with different versions of lines and manipulating equations.
2. Write the equation of any line parallel to
a) x = 3
b) y = -5
c) y = ¼x – 10
d) 2x – 3y = 5.
Equations of Lines → Why do it? Writing equations of lines parallel and perpendicular to curves is a fundamental idea →
students must be comfortable with different versions of lines and manipulating equations.
3. Write the equation of any line perpendicular to
a) y = -2
b) x = 10
c) 3x – 2y = 8
d) y – 3 = -2(x + 1)
Equations of Lines → Why do it? Writing equations of lines parallel and perpendicular to curves is a fundamental idea →
students must be comfortable with different versions of lines and manipulating equations.
4. Write the equation of a line
a) parallel to 5y + 2x = 9 and passing through (-1, 3).
b) perpendicular to y = ¼x – 8 and passing through (3, -4).
c) with slope undefined and passing through (-2, -1).
d) with slope, 0, and passing through (3, -5).
e) with x-Intercept of 2 and y-Intercept of -4.
Equations of Lines → Why do it? Writing equations of lines parallel and perpendicular to curves is a fundamental idea →
students must be comfortable with different versions of lines and manipulating equations.
5a) Draw a rough acute triangle ABC with vertices: A(-2, 5), B(4, -1) and C(0, -7). Find the equation of the median AX
where X lies on BC. Hint! The median of a triangle is defined as the segment joining a vertex to the mid-point of the
opposite side. OR it is the segment joining the mid-point of a side to the vertex opposite to it.
b) Draw a rough acute triangle ABC with vertices: A(-2, 5), B(4, -1) and C(0, -7). Find the equation of the perpendicular
bisector through side AB. Hint: The perpendicular bisector of a triangle is defined as the line that bisects any side and
perpendicular to it. [Note: The perpendicular bisector need not pass through the opposite vertex!]
c) Draw a rough acute triangle ABC with vertices: A(-2, 5), B(4, -1) and C(0, -7). Find the equation of the altitude to side
AC [Tip! The altitude shall pass through B.] Hint! The altitude of a triangle is defined as the perpendicular dropped from a
vertex onto the opposite side. OR it is the line perpendicular to a side passing through the vertex opposite to it.
Operations and Composition of functions → Why do it? Basic operations on functions is critical to calculus.
6. Given f(x) = 3x2 – 2x + 1 and g(x) = 2x – 4, find
a) f(g(x))
b) g(f(x))
c) g(g(x))
d) f(f(-1))
e) f(x) + g(x)
f) g(x) – f(x)
g) f(x)·g(x)
Division of Polynomial → Why do it? This is helpful in determining end-behaviour of functions.
7. Divide f(x) by g(x) by Long Division or Synthetic Division:
a) f(x) = 4x3 – 7x2 – 11x + 5, g(x) = 4x + 5
b) f(x) = 6x3 + 10x2 + x + 8, g(x) = 2x2 + 1
c) f(x) = x4 + 5x3 + 6x2 – x – 2, g(x) = x + 2
Solving Logarithmic and Exponential Equations → Why do it? They help determine x-intercepts of the underlying
functions.
8. Solve the following equations:
a) e-3x = 4
b) ln (x + 1) = 3
c) 25x – 2 = 17
d) 10x + 1 = 5
Expanding Expressions using Logarithms → Why do it? Application of basic log rules and properties is an integral
component of calculus.
9. Expand using Properties of Logarithms:
a) ln [(5x2 ∙ √y)/(4w · 3√z)]
b) log [x3∙(x2 – 1)]
c) log (1 – x)1/x
d) ln [(x – 2)4/(y + 2)5]
e) log x2(x + 1)3(x + 3)4/3(x – 2)4
Exploring the intersection of curves → Why do it? Behaviours of multiple functions in a plane is an important idea in
calculus.
10. Solve the system of equations:
a) 2x + y = 9;
3x – 4y = 8
b) y = -3x – 1;
4x + 5y = 6
2
c) y = x – 5x + 6
y = -7x + 5
d) y = x – 2x – 3;
y = -x + 3
2
e) y = 2x – 4x + 7
y = -x -5
2
f) y = -x + 5x;
y = x2 + 3x – 4
g) y = -x2 + 7x – 6;
y = x2 + x – 6
2
Difference Quotient → Why do it? The expression motivates the slope of the secant – and later – tangent line to a graph.
11a) For f(x) = -3x + 5, calculate f(1 + h), then f(1), then evaluate [f(1 + h) – f(1)]/h
b) For f(x) = x2 – 2x + 9, calculate f(2 + h), then f(2), then evaluate [f(2 + h) – f(2)]/h
Transformations → Why do it? Students ought to have high level of comfort with different “families” of graphs and
know the effects of simple transformations.
12. Sketch
f(x) = x2,
g(x) = √x,
c(x) = cos x,
t(x) = √x,
3
h(x) = |x|,
r(x) = x3,
s(x) = sin x,
x
u(x) = e .
Then sketch
a) y = f(x – 2)
b) y = g(x + 3)
c) y = 5h(x)
d) y = ¼r(x)
e) y = s(x – π)
f) y = c(x + π/2)
g) y = -t(x)
h) y = g(-x)
i) y = |s(x)|
j) y = u(x) + 3
k) y = c(x) – 1
Difference Quotient → Why do it? The expression motivates the slope of the secant – and later – tangent line to a graph.
13. For the functions below, find f(x + h) 1st, then simplify the expression: [f(x + h) – f(x)] / h
a) f(x) = 3x – 1
b) f(x) = 2x2 – 3x + 5
Difference Quotient → Why do it? The expression motivates the slope of the secant – and later – tangent line to a graph.
14a) If f(x) = 3x – 2, simplify the expression [f(x) – f(1)] / (x – 1).
b) If f(x) = 2x2 – 4x + 5, simplify the expression [f(x) – f(-1)] / (x + 1).
Average Rate of Change → Why do it? The expression motivates the slope of the secant line to a graph.
15. Find the Average Rate of Change of f(x) = 2x2 – 3x + 1 from x = -2 to x = 3.
Symmetricity → Why do it? This helps in graphing functions as well as suggests short-cuts for area under the curve.
16. Check if the functions below are symmetric about the x-axis, y-axis or the origin:
a) x2/16 + y2/9 = 1
b) y = -3x3/(x2 + 1)
c) y = x/|x|
d) y = sin x
e) y = cos x
f) y = tan x
Intercepts of graphs → Why do it? Students ought be comfortable with various components of graphs.
17. Find the x- and y-intercepts for the following equations [you shouldn’t use a calculator!]:
a) 2x – 3y = 5
b) y = -2(x + 1)2 + 8
c) y = 3(x – 1)3 + 81
d) y = ½|x – 4| – 6
e) y = ¼(x – 3)(x + 8)
g) y = 2x2 – x – 3
h) y = (3x2 – 5x – 2) / (2x2 + x – 3)
i) y = (x2 + 4)/x
j) y = -5/ (x2 + 1)
k) y = ln (x – 3)
l) y = ½ log(x + 10) – 3
m) y = -2ex – 1 + 4
n) y = 3 (½)x – 3 – 12
o) y = 2x3 – 3x2 + 4x – 6
p) y = 4x2 – 49
q) y = -3x2 + 27
r) y = -3x2 + 27x
s) y = 2x2 – 6x – 5
t) x2 + y2 + 7x + 2y – 8 = 0.
Graphing functions → Why do it? Curve-tracing is a critical component of calculus.
18. Graph the following functions:
Expectations: you should not plot points to graph the functions; indeed, find as few of the following as you want [!] to get
an excellent idea of what the function shall look like and use your knowledge about the general shape of the functions to
graph them quickly!

Vertex [or “Critical Point”],

the x-intercepts and y-intercepts,

Vertical and Horizontal asymptotes
You do not need to find ALL of the above. You just need to pick and choose as few or many components to help you graph!
For example,

for y = 2x2 + 4x – 3, knowing only the Vertex via (x = -b/2a, y = f(x)) and the y-intercept shall produce a good
graph. The x-intercepts are tedious to find [Quadratic formula anyone!], and asymptotes are irrelevant → draw the
parabola.

for y = 2(x + 1)3 – 16, we can find the Vertex / Critical Point (-1, -16) and the y-Intercept (0, -14) → draw a snake
up graph through those points!

For y = -e – x → y = -e – x + 0, knowing the general shape of the exponential decay function with the horizontal
asymptote of y = 0 is adequate → flip the graph across the x-axis because of the negative sign indicating reflection.
a) 2x – 3y = 5
b) y = -2(x + 1)2 + 8
c) y = 3(x – 1)3 + 81
d) y = ½|x – 4| – 6
e) y = ¼(x – 3)(x + 8)
f) y = 3x – 4
g) y = 2x2 – x – 3
h) y = (3x2 – 5x – 2) / (2x2 + x – 3)
i) y = x2/ (x2 – 9)
j) y = -5/(x2 + 1)
k) y = 4x
l) y = -log(x)
m) y = –ex
n) y = ln x
o) y = 2x3 – 3x2 + 4x – 6
p) y = 4x2 – 49
q) y = -3x2 + 27
r) y = -3x2 + 27x
s) y = e-x
t) y = (¾)x
u) y = ln (x – 1)
v) y = 3|x| + 4
w) y = ln (x – 3)
x) y = 4x2 – 4x + 1
Domain and Range of functions → Why do it? Because components of equations and graphs and “restrictions” is a
critical component of calculus.
19. Use the graphs in #18 to state the Domain and Range of each function.
Domain and Range of functions → Why do it? Because components of equations and graphs and “restrictions” is a
critical component of calculus.
20. Write the Domain of the following in Interval Notation [the "/" symbol is the division sign and whatever follows denotes
the denominator!]:
a) f(x) = (x + 2) / (2x – 7)
b) f(x) = √(2x – 5)
c) f(x) = (2x – 3) / (x2 + 4)
d) f(x) = (x – 4) / √(3x + 4)
e) g(x) = 6x / (x2 – 3x)
f) f(x) = -log(x – 1) + 3
g) f(x) = ex – 5
Graphing functions → Why do it? Curve-tracing of rational functions is a critical component of calculus.
21. Find the x-Intercept, y-Intercept, the Vertical Asymptote and Horizontal Asymptote and use test-values and graph:
a) y = 1/x
b) y = -2/x2
c) y = -4/x
d) y = 5/x2
e) y = 3/ (x – 4)
f) y = 5/ (x – 3)2
g) y = 6x/ (x – 3)
h) y = -2x/ (x + 1)
i) y = 4x/ (x + 2)2
j) y = 9x/(x + 4)2
k) y = (2x – 4) /(x + 6)
l) y = (x2 – 9) /(x2 – 4)
m) y = 3x/(x2 – 5x + 6)
n) y = 9x/(x2 – 16)
o) y = (3x + 9) /(x2 – 25)
p) y = 5x2/(x2 – 9)
Factoring → Why do it? These are basic skills that are helpful in graphing.
22. Factor the following expressions completely:
a) x3 – 8
b) 27x3 + 1
c) 3m3 – m2 + 9m – 3
d) 16x3 – 44x2 – 42x
Solving Inequalities → Why do it? The skill is critical for finding Extrema of functions and Inflection Points in calculus.
23. Solve the inequalities:
a) x2 – 5x – 6 < 0
b) x2 < 4 Careful!
c) (x + 5)(x – 2)(x + 3) > 0
d) (x + 1)(x – 2)2 < 0
e) (x – 2)/(x + 1) < 0 Careful!
f) (x – 2)/[(x – 3)(x + 4)2] > 0
g) 2x2 – x – 6 > 0
Evaluating simple logarithmic values → Why do it? This is a requirement in finding points of Extrema for trigonometric
functions in Calculus.
Finding the Inverse of functions → Why do it?
24. Evaluate without using a calculator:
a) log3 27
b) log½(4)
c) ln e2
d) log 0.1
e) log 10
f) log3 (1/27)
g) log5 1
Finding the Inverse of functions → Why do it? This is a requirement in finding the Derivatives of Inverse functions.
25. Find the inverse function of:
a) y = x – 2
b) y = (x – 2)
c) y = 2x – 3
d) y = ln(x – 4)
e) y = 10x – 3
f) y = ex – 7
g) y = log (2x – 5) + 7
h) y = 3x + 1 + 7
Converting Degress to Radians and vice versa → Why do it? This is a basic skill.
26. Convert to radians:
a) 240°
b) 150°
c) 180°
d) 330°
e) 210°
f) 225°
Convert to degrees:
g) 2π/3
h) 7π/6
i) 3π/2
j) 5π/4
k) π
l) π/6
Finding the value of simply Trigonometric ratios → Why do it? This is a requirement in finding points of Extrema for
trigonometric functions in Calculus.
27. Find the value of
a) cos(5π/3)
b) tan (11π/6)
c) sin(3π/2)
d) cos(7π/4)
e) tan (7π/6)
Solving Trigonometric Equations → Why do it? The exercise is equivalent to finding the x-Intercept of corresponding
equations, a fundamental skill in Calculus.
28. Solve the following trigonometric equations in [0, 2 π)
a) sin x = 1
b) cos x = -1
c) tan x = 0
d) cos x = 0
e) sin x = -½
f) cos x = 3/2
g) tan x = -1
h) sin x = -1
i) tan x = -3
Graphs of Trigonometric functions → Why do it? Students ought to have a sound understanding of the properties of
basic trigonometric functions in Calculus.
29. Sketch the following functions over 1 cycle (period):
a) y = sin x
b) y = cos x
c) y = -2 sin x
d) y = -cos 2x
e) y = 3 sin (x/2)
f) y = -sin πx
g) y = -5cos 4πx