Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Page 1 of 6
Math 142 Exam I Review
Slope:
y 2  y1 y  3
means as x increases 2 units, y decreases 3 units


x 2  x1 x
2
Horizontal lines: m  0
m
Vertical lines: slope is undefined
Linear Functions: straight lines that are not vertical
1. C  2.5 x  1000
Since C(0)=1000, the fixed costs are 1000.
The variable costs are 2.5 per item.
C ( x)  f  v  x
2. If given two pair of points and asked for a linear function
input data in L1 and L2
change window to match data, entering the x – data into L1 , and the y- data into L2
stat, -> to calculate, 4:LinReg L1 , L2 , y1 enter
3. supply and demand
Supply equations: positive slope
Demand equations: negative slope
Intersection: Equilibrium point (x,p)
x is the equilibrium quantity and
p is the equilibrium price
Functions: each x is paired with exactly one y response.
Domain: possible values for x is all reals unless:
a) fraction : check to see if any x values make the denominator 0.
b) even-index radical a  , if the root a is even, then M must be non-negative M  0
c) logarithm: log (a), a  0 the argument must be positive.
d) word problems – only values of x which make sense are allowed.
Vertical line test for functions
Horizontal line test for 1-1
A function must be 1-1 to have an inverse.
Find the inverse by switching all x’s and y’s and solve for the new y; rename y, f
1
( x)
Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Special Functions:
Power functions
Page 2 of 6
f ( x)  k  x r
y  4 x3
Polynomial functions f ( x)  an x n  an 1 x n 1   a1 x  a0
coefficients are real and exponents are non-negative integers
Rational function
g ( x)
where g ( x) and h( x ) are polynomials
h( x )
 x, x  0
f ( x)  x  
  x, x  0
f ( x) 
Absolute Value Function:
Application Problems
Rx p
P R C
Vertical and Horizontal shifts
f ( x)  ( x  a) 5  6
rigid transformation of the graph of y  x 5 , shifted a units right, and b units up.
Reflections
f ( x)   x 5
the graph of y  x 5 reflected about the x-axis
Stretches and Shrinks
f ( x )  6 x 5 similar to the graph of y  x 5 , stretched vertically by a factor of 6
f ( x) 
1 5
x similar to the graph of y  x 5 , vertical shrink by a factor of ½
2
Combinations: shifts up or down must occur AFTER reflections about the x-axis.
Quadratic Functions:
Standard form:
y  ax 2  bx  c , a  0
Vertex form: y  a( x  h) 2  k where (h, k) is the vertex.
Where is the vertex?
Where is the axis of symmetry?
What is the maximum value?
Find the vertex of f ( x )  5 x 2  40 x  300
 (40)
h
 4
2(5)
b
A) By formula h 
2a
b2
k c
4a
 40 
k  300 
4  5 
k  300   80   380
2
or k  f ( h)  5(4)  40( 4)  300  380
2
so the vertex is at (-4, 380)
Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Page 3 of 6
B) By completing the square:
f ( x )  5 x 2  40 x  300
f ( x )  5( x 2  8 x )  300
f ( x )  5( x 2  8 x  16)  300  80
f ( x )  5( x  4) 2  380
therefore the vertex is at (-4,380)
C) by calculator
y1  5 x 2  40 x  300 , 2nd calc maximum
Note: this is often an approximate answer and not exact!
Quadratic formula
If ax 2  bx  c  0, a  0, then x 
 b  b 2  4ac
if b 2  4ac  0 .
2a
Intercepts:
Substitute in x  0 , and solve for y
Substitute in y  0, and solve for x.
Line of symmetry passes through the vertex
x=h
Max or Min value is k depending on if the
Parabola opens up or down.
Break-Even Analysis:
R  C or P  0
Polynomial Functions:
y  ax n  bx n 1  cx n  2    z
where all the exponents are non-negative integers
Rational Functions:
f ( x) 
n( x )
d ( x)
where n(x) and d(x) are polynomials
Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Page 4 of 6
Exponential and Logarithmic Functions:
f ( x )  a  b x , b  0, b  1
y  log b x  b y  x
y  a x is increasing and concave up if a  1 and
Is decreasing and concave up if 0  a  1
x
y
a  a  x  y if a  1
When 0  a  1 , the graph approaches the x-axis as x  
When a  1 , the graph approaches the x-axis as x  
1
1
, a x  x
a
a
a 0  1, a 1 
a
1
2
 a, a
1
3
 3 a, a
1
n
na
a x  a y  a x y
ax
 a x y
y
a
a 
x
y
 a xy
Logarithms:
a a    x if x  0
log a a x  x
log
x
log a xy  log a x  log a y
log a
x
 log a x  log a y
y
log a x c  c  log a x
y  ln x iff e y  x , and log x  y iff 10 y  x
log b ln b
Change of base formula: log a b 

log a ln a
Important:
Limits:
Notation for limits from the left or right:
lim f ( x)
x 5 
lim f ( x )
x 5 
To find a limit at a constant, say 5, plug in 5. If a zero occurs in the denominator, factor, reduce and
try plugging in 5 again.
If from the left and right the graph is approaching infinity, then the limit is infinity.
If from the left and right the graph is approaching negative infinity, then the limit is negative infinity.
Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Page 5 of 6
If from the left, the graph is approaching infinity, and from the right is approaching negative infinity,
then the limit DNE (does not exist).
If from the left and right the graph is approaching two different y values, then the limit does not exist.
Continuity:
A function is continuous at x = a if all three of the following conditions are met.
1. The limit of the function exists at x = a.
2. The function is defined when x = a.
3. The limit of the function as x -> a, is the same value as the function evaluated at a.
Difference Quotient:
f ( x  h)  f ( x )

h
To find the difference quotient for any function f ( x ) ,
a. find f ( x + h )
b. find f ( x + h ) – f ( x )
c. divide by h
f ( x) 
Asymptotes:
ax 4  x3  x 2  x  8
bx 4  2 x 3  5 x  2
Horizontal Asymptotes
Compare the degrees of the numerator and the denominator
a. if the same, then there is a horizontal asymptote at y 
a
,
b
the quotient of the leading coefficients
b. if different, determine which has the larger degree
1. if the numerator is of larger degree, there is NO horizontal asymptote
2. if the denominator is of larger degree, then y  0 is a horizontal asymptote
Holes in the Graph
If given a rational function, factor and see if any of the factors cancel. Set any factor that
cancels, equal to zero, and you have the location of a hole in the graph.
Vertical Asymptote: After canceling and reducing, and remaining factors can be set equal to
zero and solved, and you have the line, x = a , which is a vertical asymptote. There may
be more than one vertical asymptote.
Limits at Inifinity: finding horizontal asymptotes
End Line Behavior: answering the questions:
a. As x  , what is happening to y? or
b. As x  , what is happening to y?
Find the degree of the polynomial.
Math 142 Lecture Notes - © Drost 9/21/2014
Section A-8 through 3-3
Page 6 of 6
a. If even, both ends of the line head in the same direction as the sign of the leading
coefficient. Plus = up and Negative = down
b. If odd, both ends of the line head in opposite directions, one to infinity, and the other
to negative infinity. To determine the direction, look at the sign of the leading
coefficient.
Interest Rates:
mt


r

m
Continuously Compounded: A  Per t
Compound Interest: F  P  1 
m
r

Effective Yield reff   1    1 OR can be calculated on TVM Solver on calculator
 m
If compounded m times a year
Effective Yield on continuously compounded account
reff  er  1
Cost, Revenue, and Profit:
C ( x)  f  v  x
R( x)  p  x
P  R C
Average Rate of Change: means slope m 
Finding the equation of the tangent line
y2  y1
x2  x1
y  y1  m( x  x1 ) Substitute in  x1 , y1  , a point on the
line, and m = slope.
Instantaneous Rate of Change:
Find the derivative by the limit definition
f '  x   lim
h 0
f ( x  h)  f ( x )
h
m  f '( x1 )
When the Derivative Fails to Exist: a corner or cusp, a hole, a jump or a vertical tangent