Download PRECALCULUS Summer Review Assignments – Equation and

PRECALCULUS Summer Review Assignments – Equation and Inequality Solving
To be prepared for your subsequent math courses, you need to be able to accurately and efficiently solve any
linear equation or inequality. This includes both absolute value sentences, and also any associated word
problems. As you will see at several times in mathematics, you will also ALWAYS need to check your work.
For problems 1 – 6, solve each equation and be sure to check your solutions.
1.
9x – 6 = 3(3x – 2)
2.
3.
9x – 2 – 3x = 2(3x – 1)
4.
5.
5x 8
2
6.
3
7
2x – 5 = -9
3
2
5
5
x
8
x
5
4
6
5(2x – 7) = 6x – 9 + 4x
NOTE:
The FAMOUS BONGIOVANNI THEOREM SAYS that if there is more than one unknown quantity
mentioned in a problem, then the second quantity mentioned is the only VARIABLE in that prompt, and the other
unknown quantities mentioned need to be expressed in terms of that variable. We will use this theorem in the following
problems. In each problem, define a variable, write expressions for each quantity, and then write and solve an equation.
EXAMPLE: The number of shirts sold was 18 more than the number of pants. If 240 total pieces of clothing were
sold, determine the number of shirts that were sold.
Define a variable:
x = the number of pants sold
Expression
x + 18 = the number of shirts sold
Expression
240 = the total pieces of clothing sold
Equation
x + x + 18 = 240
For problems 7 – 10, use the FAMOUS BONGIOVANNI METHOD to set up and solve an equation.
7.
The length of a rectangular room is 9 more than twice the width. The perimeter is 48. Find the length and width
of the room.
8.
Guido is 7 years older than Rocco. Luigi is twice as old as Guido. The sum of their ages is 61. Find the age of
each person.
9.
The perimeter of a parallelogram is 28 inches. What is the length of the shorter side if the longer side measures
8.6 inches?
10.
Twenty-three less than five times some number is 97. What is the number?
NOTE:
When graphing an inequality on a number line, when the inequality is < or >, use a open dot, and when
the inequality is < or >, use a closed dot. When solving an inequality, if you multiply or divide both sides of the
inequality by a negative, you must change the direction of the inequality sign!!
For problems 11 – 18, solve each equation and graph your solution on a number line.
11.
5 – 7x > -30
12.
5 + 3x < 2( 5x – 6 )
13.
11x + 17 > -2
14.
6x + 10 > 8 – (x + 14)
15.
w + 5 < 2 – (3w – 4)
16.
-7x – (3x + 6) > 4
17.
2( 5x + 3) + 5 > 10x + 11
NOTE:
With absolute value equations, there are a couple things to keep in mind. The first thing that needs to
happen is to isolate the absolute value quantity on one side of the equation. From there, we rewrite two separate
sentences that would look like this:
│something │ = #
IF the ORIGINAL absolute value equation is
THEN the two NEW equations would be
something = #
-1(something) = #
When each sentence has been solved, we MUST remember to check each of the solutions in the original equation. We
do this to look for extraneous solutions, which are solutions that satisfy a transformed equation, but not the original
equation. An absolute value equation will have no solution when the absolute value quantity is set equal to a negative
number.
For problems 19 – 24, solve each of the absolute value equations; be sure to check for extraneous solutions.
18.
5
x+3
20.
20 +
22.
2 2x + 4
= 25
7y + 3
= 19
19.
5/6 y + 2
=8
21.
-6 – 2 2x – 4
= -8
= 6x – 7
NOTE:
Two words that describe a conjunction are AND as well as INTERSECTION. The graph of a
conjunction would include the values that are included in each of the solution sets. Two words that describe a
disjunction are OR as well as UNION. The graph of a disjunction would include the values that are included in either of
the solution sets. Conjunctions and disjunctions are both examples of compound inequalities.
For problems 25 – 32, graph the solutions to the conjunctions and disjunctions.
23.
x > -6 and x < -3
24.
x > 5 or x ≤ 3
25.
7≤x<4
26.
x >3 or x ≤ 7
27.
x ≥ -5 and x > -2
28.
x > 4 or x > 7
For problems 33 and 34, solve the absolute value inequality, and graph the solution set.
29.
5x – 15 < 10
30.
-6 + 2 4x – 4 > 22
PRECALCULUS Summer Review Assignments – Graphing
To be prepared for algebra two, and all subsequent math courses, you need to be able to accurately and efficiently
find the slope of any line, be able to write the equation of a line given some basic information, and also be able to
graph a line when given some basic information, or an equation in any form.
1.
What is slope-intercept formula? (y = mx + b). In this formula, m stands for slope, b stands for the
y-intercept, and (x, y) stands for any point on the line.
2.
What is point-slope formula? [ y – y1 = m(x – x1) ] In this formula, m stands for the slope and (x1, y1) stands for
a specific point on the line, while (x, y) stands for a generic point on the line.
3.
All horizontal lines have equations in the form of y = a constant value
4.
All vertical lines have equations in the form of x = a constant value
5.
The x-intercept is the point where the graph intersects the x-axis, and is the point where y equals 0.
6.
The y-intercept is the point where the graph intersects the y-axis, and is the point where x equals 0.
Note: Finding both the x and y intercept can allow us to graph a linear equation very quickly since it is an application of
using the number 0. If we can find both the x and intercepts quickly, we can plot them, connect with a line, and finish
the graph, making sure to label all important features!
7.
Slope can be described in many different ways. In equations, we always represent slope by the variable m. Slope
is seen to be the steepness of the line. The slope of a line tells which direction it rises or falls. There are four different
types of slope: positive, negative, undefined, and zero.
8.
The formula for slope when given two points is: m = y2 – y1/x2 – x1
which tells us that slope is constant and can be described using the phrase RISE over RUN.
Note: If we are asked to graph an equation and are given the slope and a point on the graph, we can now do this without
solving for the x and y intercepts and without the need to make an x/y table. We would start our graph at the y intercept
and then move to other points on the graph by using the slope be thinking about rise over run. Finally we need to
complete the graph by connecting the points with a line, making sure to label all important features!
For exercises 11 – 13, complete your graphs by finding the x and y intercepts first and then using that information
to graph the lines ON GRAPH PAPER. Be sure to label the x and y axis, the key ordered pairs of the graph, and
to put arrows on both the x and y axis and the ends of the graphed line itself.
1.
2y + 3x = -18
2.
2x = 5
3.
-5y = 11
x-intercept = __________
x-intercept = __________
x-intercept = __________
y-intercept = __________
y-intercept = __________
y-intercept = __________
4.
Given a line with a slope of -2/3, determine the slope of a line that is parallel to this line. __________
5.
Given a line with a slope of 5, determine the slope of a line that is perpendicular to this line. ______
For exercises 16 – 21, graph each equation on a separate piece of graph paper, using any method.
6.
2y + 3x = -18
7.
2x = 5
8.
-5y = 11
For exercises 22 – 29, graph each line remembering special facts about parallel and perpendicular lines’ slopes.
Be sure to label the x and y axis, the key ordered pairs of the graph, and to put arrows on both the x and y axis
and the ends of the graphed line itself.
9.
10.
11.
12.
Graph a line passing through (-4, -2) that is parallel to a line with a slope of -3.
Graph a line passing through (-2, -1) that is parallel to the line with equation x = 7.
Graph a line passing through (-9, 2) that is perpendicular to a line with a slope of 3/2.
Graph a line passing through (-7, -4) that is perpendicular to a line with an equation of y = -2
NOTES:
variable:
For an absolute value equation written in vertex form, y = a|x – h| + k, we learn the following from each
a) tells us if the graph opens up or down, and if it is wide or narrow.
The graph opens up if a is positive, and opens down if a is negative. The graph has a wide opening if the value of a is in
between -1 and 1, but not equal to 0 and has a narrow opening if the value of a is greater than 1 or less than -1.
h) tells us how far to move the vertex left or right. Remember, the movement caused by h is the opposite of what we
would expect.
k) tells us how far to move up or down, and this movement is exactly as it appears.
For exercises 30 – 38, graph each equation. Show at least five ordered pairs for each equation, making an x/y
table of points as needed.
13.
y= x
________
16.
y = x-2
19.
y = 1/3 x
________
________
14.
y=- x
17.
y = -3 x
20.
y=2 x
________
________
________
15.
y = x + 1 ________
18.
y = x - 5 ________
21.
y = x+3
________
NOTES: An asymptote is a line that a graph approaches a line but never touches the line.
To graph a rational algebraic function, we should make an x/y table. On this table, we should be sure to pick values
that approach the vertical asymptote from both the left and the right, including decimal values. We also want to pick
values of x that are very large and very small to determine the location of the horizontal asymptote. Looking at the
horizontal and vertical asymptotes, will allow us to determine the domain and range of the function.
For each function listed in exercises 39 – 42, determine the vertical asymptote, make an x/y table with at least 14
ordered pairs, graph the function, and determine the equation of the horizontal asymptote. Then, identify the
domain and range of the function.
22.
f(x) =
4
x 3
23.
f(x) =
x
x 1
24.
g(x) =
x 3
x 2
25.
y=
2x
x 5
NOTES:
One way to graph an exponential function is to make an x/y table. As part of this, we should plug in
positive and negative values, as well as 0. The important thing to look for is where the values of y all approach a
common value. This tells us where the horizontal asymptote is located.
For each function listed in exercises 43 – 46, make an x/y table with at least 10 ordered pairs, graph the function,
and determine the equation of the horizontal asymptote.
26.
f(x) = 3x
27.
f(x) =(-2)x
28.
g(x) = 4x + 2
29.
y = (½)x – 5
PRECALCULUS Summer Review Assignments – Quadratics
To be prepared for your subsequent math courses, you need to be able to accurately and efficiently utilize all
skills required for quadratic expressions and equations. These include the ability to factor, apply the quadratic
formula, and graph parabolas when the equation is written in either vertex or standard form.
1.
What is standard form of a quadratic equation?
y = ax2 + bx + c
2.
What is vertex form of a quadratic equation?
y – k = a(x – h)2 or y = a(x – h)2 + k
NOTES:
We can now solve quadratic equations is by factoring. To use this method, we must use the zero
product property, which says that for any real numbers a and b, if a*b = 0, then either a=0, b=0, or BOTH a AND b
each = 0 So, in order to solve an equation by factoring, the equation must be first set equal to 0! When we have
factored, we can find the solutions of the equation by setting each of the factors equal to 0 and then solving each of those
little equations for the variable. We can determine the maximum number of solutions of an equation by looking for the
greatest exponent of a variable. Remember, in ANY factoring problem, always start by factoring out the GCF first.
In a factoring problem, if the lead term is a negative, we should distribute -1.
For problems 33 – 38, solve each of the equations by factoring.
3.
x2 – 25 = 0
4.
7x2 – 42x = 0
5.
x2 – 10x – 24 = 0
6.
3x2 – 25x – 18 = 0
7.
x2 + 8x + 21 = 9
8.
3x3 – 48x = 0
9.
x2 + 5x – 24 = 0
10.
5x3 – 45x = 0
11.
2x2 – x – 15 = 0
12.
x2 + 6x + 8 = 0
13.
What is the Quadratic Formula?
14.
What is the discriminant?
___________________________
_______________________________
NOTES:
The discriminant can tell us about the number and type of solutions of a quadratic equation. There are
four cases of the discriminant. They are:
A negative discriminant has TWO imaginary solutions
A discriminant that is positive and a perfect square has TWO real, rational solutions
A discriminant that is positive, but not a perfect square has TWO real, itrational solutions
A discriminant of zero has ONE real, rational solution
For problems 45 – 56, a) determine the discriminant, b) determine the specific type and number of solutions, and
c) solve the equation using the quadratic formula. Leave all answers in simplest radical (or imaginary) form.
15.
x2 – 10x + 24 = 0
16.
2x2 – 10x – 5 = 0
17.
2x2 – 12x + 72 = 0
18.
5x2 – 45 = 0
19.
x2 + 3x = 28
20.
x2 + 9x + 31 = 9
NOTES:
A parabola is the graph of a quadratic function. It is in the shape of a U.
To graph the quadratic, we can find the vertex, the axis of symmetry, and the x and y intercepts.
The vertex is point of intersection of the parabola and axis of symmetry. It is also the high or low point of the graph.
b
To find the vertex, first find its x-coordinate by evaluating
when the equation is written in standard form. Then, to
2a
find the y-coordinate of the vertex, plug the x-coordinate back into the original equation. To find the y-intercept of the
equation, find the value of C when the equation is written in standard form; so it is (0, C). The axis of symmetry is line
about which the graph is symmetric. The x-coordinate of the vertex also tells us the line of symmetry of the
equation. The equation of the line of symmetry is the vertical line that goes through the vertex. Remember, all vertical
b
lines are in the form of x = “a number”, so the equation of the axis of symmetry is x =
. We can use the line of
2a
symmetry to find a point on the graph symmetrical to the y-intercept.
The roots (or solutions) of a quadratic equation are values of the variable that satisfy the equation. These are places at
which the graph crosses the x-axis, so these are the x-intercepts and are therefore the points where y = 0. These are also
known as the zero’s of the function. To find these, solve the equation by using either the quadratic formula, or by
factoring. Then, after evaluating one more point, and finding its symmetrical point, the graph can be sketched.
For a parabolic equation written in vertex form, y = a(x – h)2 + k, we learn the following from each variable:
a) tells us if the graph opens up or down, and if it is wide or narrow
the graph opens up if a is positive, and opens down if a is negative
the graph has a wide opening if the value of a is in between 1 and -1, but not equal to 0 and has a narrow opening if the
value of a is greater than 1 or less than -1
h) tells us how far to move the vertex left or right. Remember, the movement caused by h is the opposite of what we
would expect.
k) tells us how far to move up and down, and this movement is exactly as it appears.
For the following problems, a) determine the vertex and equation of the axis of symmetry, b) determine the x- and
y-intercepts, if they exist, and c) graph the quadratic showing at least five ordered pairs on the graph.
21.
y = x2 – 4x + 4
22.
y = -2x2 - 2x + 3
23.
f(x) = x2 – 4x – 12
24.
y = -.5x2 – 8x + 10
25.
y = x2 + 3
26.
y = 5x2
27.
y = -½x2
28.
y = (x – 2)2 + 3
29.
y = -2(x + 7)2 – 5
PRECALCULUS Summer Review Assignments – Radicals
To be prepared for your subsequent math courses, you need to be able to accurately and efficiently simplify a radical
statement. This includes adding, subtracting, all forms of multiplying, and dividing. As you will see at several times in
mathematics, you will also ALWAYS need to check your work when solving a radical equation.
NOTES:
In the expression, 25 the
is called the radical sign, the 25 is referred to as the radicand and the whole
expression is called a radical. When the square root of a number is taken, the two answers are called the principal and negative
square roots. In the expression, n x a , the ‘n’ is the root index. If no root index is shown, then it is assumed to be 2, and the root is
called a square root. When the root index is even, the radicand must be positive for the value to be real.
The directions to problems in this section will only say to ‘simplify.’ While that seems like a harmless little word, it carries much
weight here. A radical expression has been completely simplified, only if each of the following are all completed:
1.
All radicals that can be ‘combined’ by addition, subtraction, or multiplication have been combined.
2.
There are no perfect square factors in the radicand.
3.
There are no radicals in the denominator
4.
All fractions have been reduced.
5.
The root index is as small as possible.
Radicals are like variables. You may only multiply and divide numbers if they both have the square root, or both don’t have the
square root.
OK:
or
NOT OK:
or
The product property of radicals shows that:
This property allows us to simplify a radical expression by finding the largest perfect square that is a factor of the radicand,
separating it into two radical expressions, then simplifying the perfect root factor.
For example,
24 =
4
6 =2 6
To simplify an expression such as x 20 , we would divide the exponent of the radicand by the root index (in this case, 2. To
simplify an expression such as, 3 x 20 , divide the exponent of the radicand by the root index (in this case, 3; leave the quotient as a
power on the outside and the remainder as a power under the radical.
The rules for adding and subtracting radicals are like those for combining like terms
We can only add or subtract radicals when the radicands AND root indexes are alike.
When we add or subtract radicals, remember do NOT add or subtract the radicands OR root indexes.
Also, remember, NEVER COMBINE unlike radicals.
If a radicand is a fraction, it can be rewritten into two separate radicals.
To rationalize the denominator, we need to multiply by a value of 1. We can do this because any number times 1 equals itself.
The value can be expressed by some radical divided by itself. We will know what radicand to use by the radical in the
denominator.
When trying to rationalize a denominator in a problem such as
5
3
x7
there are three things to keep in mind about the “target
denominator” that we are trying to get. These things to remember are:
The root index needs to divide evenly into the new exponent.
The new exponent needs to be greater than the original exponent, but
The new exponent should be kept as small as possible.
Conjugate binomials are two binomials with the same terms but different signs.
(a + b )(a - b ) are an example of conjugate binomials. When we multiply them, the answer will be a rational number not
containing any radicals.
To simplify a radical expression with “two-terms” in the denominator, we need to multiply both the numerator and the
denominator by the conjugate of the denominator. For example:
The following refers to higher order radicals, or the nth root, for any real number x, and any integer n, where n is greater than one:
n
b =
, and
n
bm =
=(
)m. The ‘n’ is called the root index in either of these cases.
Remember, we may not have an expression that has been completely simplified with a fractional exponent in the denominator, as
the fractional exponent represents a radical.
Imaginary numbers are realized when the radicand of a radical expression is negative. We simplify imaginary numbers using the
same methods for any other radical except that we must start by factoring out
, which is represented by i.
x2 = -1 cannot be solved in the real number system. Rene Descartes, a mathematician from France proposed using the number i.
When rationalizing denominators, we must also eliminate i in the denominator as well, because this is a representation of
. To
simplify a radical or imaginary denominator with a binomial, we need to multiply both the numerator and the denominator by the
conjugate of the denominator.
1
-1 = i2
=i
- 1i = i3
1 = i4
We can then simplify higher powers of i by dividing the exponent of i by 4, and then look at the remainder to see what the
imaginary number would simplify to. An answer in terms of i in simplest form may not contain an exponent of i that is greater
than 1.
The sum (or difference) of a real and imaginary number is called a complex number. These are any numbers that can be written in
the form of a + bi where a is the real part and bi is the imaginary part. These numbers can be graphed on a Cartesian Coordinate
System where the real value falls along the ‘x’ axis and the imaginary value falls along the ‘y’ axis.
To add or subtract complex numbers, just follow the rules for combining like terms.
To multiply complex numbers, just follow the rules for FOIL.
A radical equation is an equation with a radical expression in one of the members.
An extraneous solution is one that satisfies a transformed equation, but not the original.
To solve a radical equation, start by isolating the radical on one side of the equation. When that is completed, then raise both
sides of the equation to the power of the root index to remove the variable from under the radical. We can then look to isolate
the variable using prior methods. When the variable has finally been isolated, we must check the possible solution in the
original equation to look for extraneous solutions.
For problems 1 – 195, simplify each radical expression completely.
1.
16
2.
169
3.
17
4.
3 25
5.
x2
6.
12
7.
72
8.
363
9.
16 y 21 z 15
10.
9
11.
± 196
12.
6x3 50x 7
14.
54
15.
4
13.
3
17.
18.
19.
21.
22.
23.
4
25.
29.
48
26.
3
30.
125x
162x 8 y11
16.
20.
3
27c 6
24.
27.
28.
31.
32.
4
(2 x 3 y ) 4
3
7
2
33.
(3 5 )(-2 2 )
34.
37.
6 ∙ 35 ∙ 10
38.
41.
2 3
2
44.
48.
5 6 3 7
8
75
54.
5
7
8
45.
4 7
6
3
3
20
108
55.
25
8
2
39.
2
43.
12 46.
47.
49.
6 7
52.
1
2
4
60.
6x
2 7
2
3
6 2 27
5 50
3 4 3
5
15
5
56.
7x3 12x15
36.
40.
42.
2 6
51.
59.
6
5 (2 3 - 4 5 )
35.
2 3
10 4 6
18
50.
58.
5
64( x 5) 2
53.
50
80
57.
3m 3
24 n 5
61.
3
6
2
3
10
For problems 196 – 243, simplify each imaginary expression completely leaving no powers of i greater than 1.
5
1
62.
63.
i2
64.
65.
3 11
3 3i
8i
3
2
66.
67.
68.
(3 + 2i) + (5 – 7i)
69.
(4 + 3i) – (4 + 7i)
4 i
70.
i20
74.
(9 – 3i) + (7 – 14i)
77.
(8i)3
81.
-2
78.
48
12 – 3i 3
84.
87.
71.
4
5i
i29
(8 + 3i)(6 – 5i)
72.
75.
7
(5i)(10i)(4i)(3i)
6
3
85.
-17i57
9
5i
2
76.
18
79.
82.
88.
75
3 +
5
80.
86.
6( 4 n )7
90.
7
5x 3 y 4
For problems 246 – 247, write each expression in radical form.
2
91.
9g3
92.
(12x)
1
2
(10 +
3 )(6 –
7)
5
3
18x10 y 7
83.
For problems 244 – 245, write each expression in exponential form.
89.
(6 – 8i) – (2 – 15i)
73.
(3
12 )(11
5 )(
3)
For problems 248 – 251, simplify each of the following expressions showing all of your work.
8
93.
2
7
y y
5
7
94.
x 11
x
5
11
95.
81
3
1
2
96.
(4) 2
For problems 252 – 265, solve each equation; be sure to check for extraneous solutions.
3x 6 – 10 = 2
97.
99.
(10x – 11)½ + 19 = 16
101.
6x 2
98.
10 x 13
100.
4
3x 8
x 3–6=3
102.
5
6n 5 = 3
4x + 3 = 8
3
103.
3
5x 7 = 6
104.
4
105.
x 5 –9=6
106.
(5x – 2)½ + 10 = 6
107.
14 + x 10 = 30
108.
13 -
109.
5 + x 6 = 17
110.
y
1
3
5x 35 = 8
5
https://www.khanacademy.org/math/algebra/exponent-equations/simplifying-radical-expressions/v/multiply-andsimplify-a-radical-expression-1