Category Archives: MAC1140

5.2 The Real Zeros of a Polynomial Function

Intermediate Value Theorem

July 5, 2018 admin

Problem: Use the Intermediate Value Theorem to show that the following function has a zero in the given interval. Approximate the zero to two decimal places.

f(x)=9x^3+9x^2-9x+6; [-2,-1]

Solution:

To determine if there is a zero in the interval use the Intermediate Value theorem. To use the Intermediate Value Theorem, the function must be continuous on the interval [-2,-1] . The function f(x)=9x^3+9x^2-9x+6 is a polynomial function and polynomial functions are defined and continuous for all real numbers.

Evaluate the function at the endpoints and if there is a sign change. If there is a sign change, the Intermediate Value Theorem states there must be a zero on the interval. To evaluate the function at the endpoints, calculate f(-2) and f(-1) .

f(-2)=9(-2)^3+9(-2)^2-9(-2)+6
f(-2)=9(-8)+9(4)+18+6
f(-2)=-72+36+24
f(-2)=-72+60
f(-2)=-12

f(-1)=9(-1)^3+9(-1)^2-9(-1)+6
f(-1)=9(-1)+9(1)+9+6
f(-1)=-9+9+9+6
f(-1)=0+15
f(-1)=15

Since one endpoint gives a negative value and one endpoint gives a positive value, there must be a zero in the interval.

We can get a better approximation of the zero by trying to figure out the next decimal point. Write out all of the values to one decimal point between -2 and -1.

x	f(x)	f(x)
-2.0	9(-2)^3+9(-2)^2-9(-2)+6	-12
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0	9(-1)^3+9(-1)^2-9(-1)+6	15

Fill the table. There are functions in your calculator that make this easier.

x	f(x)	f(x)
-2.0	9(-2)^3+9(-2)^2-9(-2)+6	-12
-1.9	9(-1.9)^3+9(-1.9)^2-9(-1.9)+6	-6.141
-1.8	9(-1.8)^3+9(-1.8)^2-9(-1.8)+6	-1.128
-1.7	9(-1.7)^3+9(-1.7)^2-9(-1.7)+6	3.093
-1.6	9(-1.6)^3+9(-1.6)^2-9(-1.6)+6	6.576
-1.5	9(-1.5)^3+9(-1.5)^2-9(-1.5)+6	9.375
-1.4	9(-1.4)^3+9(-1.4)^2-9(-1.4)+6	11.544
-1.3	9(-1.3)^3+9(-1.3)^2-9(-1.3)+6	13.137
-1.2	9(-1.2)^3+9(-1.2)^2-9(-1.2)+6	14.208
-1.1	9(-1.1)^3+9(-1.1)^2-9(-1.1)+6	14.811
-1.0	9(-1)^3+9(-1)^2-9(-1)+6	15

Use the Intermediate Value Theorem again. Look for a sign change. Looking down the table, there is a sign change between -1.8 and -1.7. With this information we now know the zero is between these two values.

Repeat this process again with two decimal places between -1.8 and -1.7.

x	f(x)	f(x)
-1.80	9(-1.8)^3+9(-1.8)^2-9(-1.8)+6	-1.128
-1.79	9(-1.79)^3+9(-1.79)^2-9(-1.79)+6	-.6712
-1.78	9(-1.78)^3+9(-1.78)^2-9(-1.78)+6	-.2222
-1.77	9(-1.77)^3+9(-1.77)^2-9(-1.77)+6	.219
-1.76	9(-1.76)^3+9(-1.76)^2-9(-1.76)+6	.65242
-1.75	9(-1.75)^3+9(-1.75)^2-9(-1.75)+6	1.0781
-1.74	9(-1.74)^3+9(-1.74)^2-9(-1.74)+6	1.4962
-1.73	9(-1.73)^3+9(-1.73)^2-9(-1.73)+6	1.9066
-1.72	9(-1.72)^3+9(-1.72)^2-9(-1.72)+6	2.3096
-1.71	9(-1.71)^3+9(-1.71)^2-9(-1.71)+6	2.705
-1.70	9(-1.7)^3+9(-1.7)^2-9(-1.7)+6	3.093

Use the Intermediate Value Theorem. Look for a sign change. Looking down the table, there is a sign change between -1.78 and -1.77. With this information we now know the zero is between these two values and the zero to two decimal places is -1.77 since all the numbers between -1.78 and -1.77 start with -1.77.

4.1 Quadratic Functions, 4.3 Quadratic Functions and Their Properties

Analyze a Quadratic Function in Vertex Form

July 3, 2018 admin

Example: Given the quadratic function in vertex form, state the domain, range, vertex, x-intercepts, y-intercept, the orientation (opens up or opens down), and the axis of symmetry. Finally graph the function.

$f(x)=-{1/4}(x-1)^2+5$

Solution:

Find the vertex.

When the quadratic function is written in standard form you can identify the vertex as (h,k).

f(x)=a(x-h)^2+k

$f(x)=-{1/4}(x-1)^2+5$

The vertex of the quadratic function is (1,5) .

Find the orientation.

The leading coefficient of the quadratic function is negative so the parabola opens down.

Find the y-intercept.

To find a y-intercept let x=0.

Start with the original function.	$f(x)=-{1/4}(x-1)^2+5$
To find a y-intercept let x=0.	$f(0)=-{1/4}(0-1)^2+5$
Simplify using order of operations. First complete operations inside parenthesis.	$f(0)=-{1/4}(0-1)^2+5$ $f(0)=-{1/4}(-1)^2+5$
Apply the exponent.	$f(0)=-{1/4}(-1)^2+5$
Multiply.
Get a common denominator and combine fractions

The y-intercept is (0,19/4) .

Find the x-intercept.

Start with the original function.	$f(x)=-{1/4}(x-1)^2+5$
To find an x-intercept let y=0 or f(x)=0.	$0=-{1/4}(x-1)^2+5$
Isolate the square by subtracting 5 on both sides. Then simplify.	$0=-{1/4}(x-1)^2+5$ $0-5=-{1/4}(x-1)^2+5-5$ $-5=-{1/4}(x-1)^2$
Isolate the square by multiplying by -4 on both sides. Then simplify.	$-5=-{1/4}(x-1)^2$ $-5(-4)=(-4)-{1/4}(x-1)^2$
Get rid of the square by square rooting both sides.	$sqrt{20}=sqrt{(x-1)^2}$
Simplify the radical by finding a perfect square factor of 20.
Isolate x by adding 1 to both sides. Then simplify.
Rewrite as two separate answers.	or

The x-intercepts are (1 + 2 sqrt{5},0) and (1 - 2 sqrt{5},0)

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is (1,5) the axis of symmetry is x=1 .

Graph the function.

Plot the intercepts and the vertex.

Since the vertex is the highest point we can draw the parabola using the peak at the vertex.

Find the domain and range.

The domain is (- infty, infty) .

The range is - infty, 5 {}{]} .

Here is a video example analyzing a quadratic function in vertex form.

4.2 Applications and Modeling of Quadratic Functions, 4.4 Build Quadratic Models from Verbal Descriptions and from Data

Application: Quadratic Minimimization

July 2, 2018 admin

Problem:

A 28 inch wire is to be cut. One piece is to be bent into the shape of a square, whereas the other piece is the bent into the shape of a rectangle whose length is twice the width. Find the width of the rectangle that will minimize the total area.

Solution:

First draw a picture of the shapes the wire will make and label the sides.

The square has equal sides. Label the side as an unknown quantity x.

The perimeter of the square is and the area of the square is x^2 .

The rectangle described as a length that is twice the width. Label the sides as y and 2y.

The perimeter of the rectangle is and the area of the rectangle is 2y^2 .

Since the 28 inch piece of wire will be cut and used to form the square and rectangle the total perimeter of the two shapes is 28 inches.

P=4x+6y=28

The total area is to be minimized.

A=2y^2+x^2

To minimize the area there must be only one variable in the expression. Use the perimeter equation to reduce the number of variables.

Perimeter Equation
Solve for one of the variables. In this example, solve for x by first subtracting 6y from both sides. Simplify.
Continue to solve for x by dividing both sides by 4 and simplifying.

Substitute into the area expression to reduce from having two variables to have one variable.

Area Expression
Substitute for x.	$A=2y^2+(7-{3/2}y)^2$
Simplify by applying the exponent and simplifying.	$A=2y^2+(7-{3/2}y)^2$ $A=2y^2+(7-{3/2}y)(7-{3/2}y)$ $A=2y^2+49-{21/3}y-{21/3}y+{9/4}y^2$ $A=49-{42/2}y+{17/4}y^2$ $A=49-21y+{17/4}y^2$

$A(y)={17/4}y^2-21y+49$ is in the form of a quadratic function. The graph of a quadratic function is a parabola. This quadratic function has a leading coefficient of 17/4 and since it is positive means that the parabola is opening up.

Find the vertex to find the minimum value.

The formula for the x coordinate of the vertex
For a quadratic function a is the coefficient of the square term, b is the coefficient of the linear term, and c is the constant	$A(y)={17/4}y^2-21y+49$
Substitute the values of a and b into the formula
Simplify with a calculator

Round to the nearest tenth and y=2.5 .

The total area of the square and rectangle is minimized when the width of the rectangle is 2.5 inches.

3.4 Transformation of Functions, 3.5 Graphing Techiques: Transformations

Graphing by Transformations: Quadratic

June 29, 2018 admin

Example: For the function below. Graph using transformations.

f(x)=-(x-3)^2+4

First we must examine the base function y=x^2

Graph using plotting points. We can use the standard set of x-values to find ordered pairs. Substitute the standard set of x-values into the base function to get the base graph.

x	y
-2	(-2)^2=4
-1	(-1)^2=1
0	(0)^2=0
1	(1)^2=1
2	(2)^2=4

The graph below shows the points plotted and the line that connects them. The domain is (- infty, infty) and the range is [ 0, infty )

Analyze the transformations.

f(x)=-(x-3)^2+4

The -3 inside the square shifts the graph right 3 units.

The – in the front of the base reflects the graph over the x-axis.

The +4 outside the square shifts the graph up 4 units.

You can see the graph after the transformations.

The domain is (- infty, infty) and the range is ( - infty, 4 ]

Here is a video example of a transformation of a square function.

3.6 One-to-one Functions; Inverse Functions, 6.2 One-to-one Functions; Inverse Functions

Verifying Two Functions are Inverses

June 26, 2018 admin

3.6 One-to-one Functions; Inverse Functions, 6.2 One-to-one Functions; Inverse Functions

Finding the Inverse of a Linear Function

June 26, 2018 admin

5.2 The Real Zeros of a Polynomial Function

Remainder Theorem and Factor Theorem

June 23, 2018 admin

Problem: Use the remainder theorem to find the remainder when f(x) is divided by x-1/5 . Then use the factor theorem to determine whether x-1/5 is a factor of f(x).

f(x)=5 x^4-x^3-5x-1

Solution:

The Remainder Theorem: Let f be a polynomial function. If f(x) is divided by x-c , then the remainder is f(c) .

The remainder theorem gives a way to find the remainder without performing the division. The remainder when dividing by x-c is f(c). Calculate f(c) to find the remainder where c is the zero of the factor x-c .