# College Algebra Tutorial 41

College Algebra
Tutorial 41: Practice Test on Tutorials 34 - 40

Problems 1a - 1b:  Find the coordinates of the vertex of the given quadratic function.  Without graphing, determine if the vertex is the maximum or minimum point of the quadratic function.

1a.

VERTEX

Since (h, k) is the vertex in standard form then the vertex is  (5, -4).

MAX OR MIN?
Since a = 1, and 1 is greater than 0, this parabola would open up.

So our vertex (5, -4) is the minimum point.

1b.

VERTEX

Using the vertex formula we get:

The vertex would be (-3, 19)

MAX OR MIN?
Since a = -1, and -1 is less than 0, this parabola would open down.

So our vertex (-3, 19) is a maximum point.

Problems 2a - 2b:  Use the vertex and the intercepts to sketch the graph of the given quadratic function.  Find the equation for this function’s axis of symmetry.

2a.

CURVE
Since a = -1 and -1 < 0, then it looks like it is going to curve down.

VERTEX

Vertex is (-1, 4)

INTERCEPTS

y-intercept:

y-intercept: (0, 3)

x-intercept:

x-intercepts: (-3, 0) and (1, 0)

GRAPH

As shown on the graph, the axis of symmetry is x = -1.

2b.

CURVE
Since a = 1 and 1 > 0, then it looks like it is going to curve up.

VERTEX

Vertex is (0, 3)

INTERCEPTS

y-intercept:

y-intercept: (0, 3)

x-intercept:

x-intercepts:
Note how we got a negative number underneath the square root.  That means there is no real number solution.  That also means that there are NO x- intercepts.

GRAPH

As shown on the graph, the axis of symmetry is x = 0.

Problem 3a:  Given the polynomial function a) use the Leading Coefficient Test to determine the graph’s end behavior, b) find the x-intercepts (or zeros) and state whether the graph crosses the x-axis or touches the x-axis and turns around at each x-intercept, c) find the y-intercept, d) determine the symmetry of the graph, e) indicate the maximum possible turning points, and f) graph.

3a.

Since the degree of the polynomial, 4,  is even and the leading coefficient, 1, is positive, then the graph of the given polynomial rises to the left and rises to the right.

b) X-INTERCEPT

First factor:

Since the exponent on this factor is 2, then multiplicity for the zero x = 0 is 2.

Since the multiplicity is 2, which is even, then the graph touches the x-axis and turns around at the zero x = 0.

Second factor:

Since the exponent on this factor is 1, then multiplicity for the zero x = -2 is 1.

Since the  multiplicity is 1, which is odd, then the graph crosses the x-axis at x = -2.

Third factor:

Since the exponent on this factor is 1, then multiplicity for the zero x = 2 is 1.

Since the  multiplicity is 1, which is odd, then the graph crosses the x-axis at x = 2.

c) Y-INTERCEPT
Letting x = 0 we get:

The y-intercept is (0, 0).

d) SYMMETRY
Y-axis symmetry:

It is symmetric about the y-axis.

Origin symmetry:

Is not symmetric about the origin.

e) TURNING POINTS
Since the degree of the function is 4, then there is at most 4 - 1 = 3 turning points.

EXTRA POINTS
To get a more accurate curve, lets find some points that are in between the points we found above:

So (-1, -3) and (1, -3) are two more points on the graph.

f) GRAPH

Problem 4a:  Divide using long division.

4a.

Problem 5a: Divide using synthetic division.

5a.

Problem 6a:  Given the function f(x), use the Remainder Theorem to find f(-2).

6a.

f(-2) = 33

Problem 7a:  Solve the given equation given that 2/3 is a zero (or root) of .

7a.

Rewriting f(x) as (x - 2/3)(quotient) we get:

Setting f(x) = 0 and solving we get:

Problems 8a - 8b:  List all of the possible zeros or roots, use Descartes’ Rule of Signs to possible narrow it down, use synthetic division to test the possible zeros or roots and find an actual zero or root, and use the actual zero to find all zeros or the actual root to find all solutions to the given polynomial function or equation to solve.

8a.

POSSIBLE ZEROS
The factors of the constant term -6 are.

The factors of the leading coefficient 1 are.

Writing the possible factors as p/q we get:

DESCARTES' RULE OF SIGNS
Possible number of positive real zeros:

The up arrows are showing where there are sign changes between successive terms, going left to right.

There is 1 sign change between successive terms, which means there is exactly one positive real zero.

Possible number of negative real zeros:

There are 2 sign changes between successive terms, which means that is the highest possible number of negative real zeros.  To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 2, and then go down by even integers from that number until you get to 1 or 0.

Since we have 2 sign changes with f(x), then there are possibility of 2 or 2 - 2 = 0 negative real zeros.

FIND A ZERO
I’m going to choose -1 to try:

We found a number that has a remainder of 0. This means that x = -1 is a zero or root of our polynomial function.

FIND OTHER ZEROS
Since, x = -1 is a zero, that means x + 1 is a factor of our polynomial function.

Rewriting f(x) as (x + 1)(quotient) we get:

We need to finish this problem by setting this equal to zero and solving it:

The zeros of this function are x =  -3, -1, and 2.

8b.

POSSIBLE ZEROS
The factors of the constant term -2 are.

The factors of the leading coefficient 1 are.

Writing the possible factors as p/q we get:

DESCARTES' RULE OF SIGNS
Possible number of positive real zeros:

The up arrows are showing where there are sign changes between successive terms, going left to right.

There is 1 sign change between successive terms, which means there is exactly one positive real zero.

Possible number of negative real zeros:

There is 1 sign change between successive terms, which means there is exactly one negative real zero.

FIND A ZERO
I’m going to choose 1 to try:

We found a number that has a remainder of 0. This means that x = 1 is a zero or root of our polynomial function.

FIND OTHER ZEROS
Since, x = 1 is a zero, that means x - 1 is a factor of our polynomial function.

Rewriting f(x) as (x - 1)(quotient) we get:

We need to finish this problem by setting this equal to zero and solving it:

The zeros of this function are x =  -1, 1, .

Problem 9a:  Show that all real roots of the given equation lie between -2 and 2.

9a.

LOWER BOUND

Since c = -2 < 0 AND the successive signs in the bottom row of our synthetic division alternate, then -2 is a lower bound for the real roots of this equation.

UPPER BOUND

Since c = 2 > 0 AND all of the  signs in the bottom row of our synthetic division are positive, then 2 is an upper bound for the real roots of this equation.

Since -2 is a lower bound and 2 is an upper bound for the real roots of the equation, then that means all real roots of the equation  lie between -2 and 2.

Problem 10a:  Show that the given polynomial has a real zero between the given integers.  Use the Intermediate Value theorem to find an approximation for this zero to the nearest tenth.

10a. ;  between 1 and 2.

Finding f(1):

Finding f(2):

Since there is a sign change between f(1) = -1 and f(2) = 10, then according to the Intermediate Value Theorem, there is at least one value between 1 and 2 that is a zero of this polynomial function.

Checking functional values at intervals of one-tenth for a sign change:

Finding f(1):

Finding f(1.1):

Finding f(1.2):

Hey we have a sign change.  Now we want to find the zero to the nearest tenth.  So is it going to be x = 1.1 or x = 1.2.   We can not necessarily go by which functional value is closer to zero.  We will need to dig a little bit deeper and go by intervals of one-hundredths.

Finding f(1.11):

Finding f(1.12):

Finding f(1.13):

Finding f(1.14):

Finding f(1.15):

Finding f(1.16):

Whew!!!!  At last we come to a sign change between successive hundredths.   There is a zero between 1.15 and 1.16.

The nearest tenth would be 1.2.

Problem 11a:  Use the given root to find all of the roots of the given polynomial equation.

11a.

The complex number 2 + i is a root, that means it’s conjugate 2 - i is also a root.

Using synthetic division with the given zero to find the quotient we get:

Using synthetic division with the conjugate 2 - i and the quotient found above we get:

From here we can rewrite the original problem using the roots that we have above and the quotient that we ended up with in this last synthetic division.

So the roots of the polynomial equation are 2 + i, 2 - i, -1/2.

Problem 12a:  Factor the given polynomial function  a) as the product of factors that are irreducible over rational numbers, b) as the product of factors that are irreducible over real numbers, and c) in completely factored form involving complex nonreal numbers.

12a.

a) FACTORS THAT ARE IRREDUCIBLE OVER RATIONAL NUMBERS

Since 7 is not a perfect square, this is as far as we can factor it using only rational numbers.

b) FACTORS THAT ARE IRREDUCIBLE OVER REAL NUMBERS

Even though 7 is not a perfect square, we can factor the first factor found above as a difference of squares and end up with an irrational number, which is still a real number.

c) COMPLETELY FACTORED OVER COMPLEX NONREAL NUMBERS

Problem 13a:  Find an nth degree polynomial function with the given conditions.

13a. n = 3;  -3i and 2 are zeros and f(1) = -20

WRITE OUT ALL FACTORS OF THE POLYNOMIAL
Since our degree is 3, that means there are three linear factors over the complex numbers (possibly real and not necessarily distinct).

Note how we are only given two zeros.  We need to come up with a third one.

Since -3i is a zero, that means its conjugate, 3i, is also a zero.

Using the Linear Factorization Theorem we get:

MULTIPLY FACTORS

FINDING a
This problem gave another condition, f(1) = -20.

PUTTING IT ALL TOGETHER

Problem 14a:  Sketch the graph of the rational function.

14a.

REDUCE

This function cannot be reduced any further.  This means that there will be no open holes on this graph.

VERTICAL ASYMPTOTES

HORIZONTAL ASYMPTOTES

SYMMETRY

Since , the function is neither even nor odd.  This means the graph is not symmetric about the y-axis nor the origin.

INTERCEPTS

y-intercept:

y-intercept: (0, 0)

x-intercept:

x-intercept: (0, 0)

GRAPH

Last revised on March 20, 2011 by Kim Seward.