Category Archives: MAC1140

Problem: Solve the exponential equation.

$36^{x}=(1/6)^{x-3}$

Solution:

The exponential equation	$36^{x}=(1/6)^{x-3}$
The bases of the exponents on each side of the equation can be made the same. 36 can be written as and can be written as $6^{-1}$	$36^{x}=(1/6)^{x-3}$ $(6^2)^{x}=(6^{-1})^{x-3}$
Use the power rule for exponents to multiply the exponents. Power Rule for Exponents $(b^m)^n=b^{mn}$	$(6^2)^{x}=(6^{-1})^{x-3}$ $6^{2x}=6^{-1(x-3)}$ $6^{2x}=6^{-x+3}$
Exponential functions are one-to-one thus giving us the property that if the bases are the same the exponents are equal. If then	$6^{2x}=6^{-x+3}$
Solve the remaining equation. This equation is linear, first get the variables to the same side.
Solve: Get the variable by itself.
Check: Substitute into the original equation.	$36^{1}=(1/6)^{1-3}$ $36=(1/6)^{-2}$ $36=6^{2}$

The solution to the equation $36^{x}=(1/6)^{x-3}$ is x=1 .

Here is a video example of a similar type of problem.

5.3 Properties of Logarithms, 6.5 Properties of Logarithms

Condense a Logarithm

April 4, 2018 admin

Problem: Use the properties of logarithms to rewrite the expression as a single logarithm. Whenever possible, evaluate logarithmic expressions.

ln sqrt{x}- 1/6 ln x +ln root{4}{x}

Soluton:

Logarithmic Expression
Use the power rule for logarithms. The coefficient of 1/6 on the middle term becomes the power on the expression inside the logarithm	$ln sqrt{x}- ln x^{1/6} +ln root{4}{x}$
A radical can be written as a fractional power. A square root is the same as the one-half power. A fourth root is the same as the one-fourth power	$ln sqrt{x}- ln x^{1/6} +ln root{4}{x}$ $ln x^{1/2}- ln x^{1/6} +ln x^{1/4}$
Condense the logarithms using the product and quotient rule.Product Rule for Logarithms: Quotient Rule for Logarithms: The expressions inside the logarithm will be positioned in the numerator if the logarithm is positive or will be positioned in the denominator if the logarithm is negative.	$ln x^{1/2}- ln x^{1/6} +ln x^{1/4}$ $ln {x^{1/2}x^{1/4}}/x^{1/6}$
Since these base of the exponential expressions are the same, combine using the power and quotient rules for exponent. Product Rule for Exponents: Quotient Rule for Exponents:	$ln {x^{1/2}x^{1/4}}/x^{1/6}$ $ln {x^(1/2+1/4)}/x^{1/6}$
Find a common denominator to combine the fractions.	$ln x^{7/12}$

Here is a video with a similar example worked out.

4.6 Rational Functions and Their Graphs, 5.4 Properties of Rational Functions, 5.5 The Graph of a Rational Function

Rational Function: Sketch the graph of the rational function

April 2, 2018 admin

Problem:

Sketch the graph of the rational function using algebra techniques.

$f(x)= 1/{x^2-9}$

Solution:

To start sketching the graph, gather the following information about the graph.

Domain, vertical asymptotes, holes/removable discontinuities, horizontal asymptotes, oblique/slant asymptotes, points where the graph crosses the horizontal asymptotes, x-intercepts, and y-intercepts.

Find the Domain of a Rational Function

Division by zero is undefined. Having a zero as the denominator is equivalent to division by zero thus is also undefined. The rational function is undefined for any value of the variable that gives a zero denominator. Find these values by creating an equation to solve. The equation is the expression in the denominator equal to zero.

Set the denominator equal to zero.
Solve the equation. This equation is a quadratic equation and can be solved by factoring, the square root method, completing the square or the quadratic formula.
Solve by the square root method. Start by isolating the square.
Cancel out the square by square rooting both sides.	$sqrt{x^2}=sqrt{9}$
The solutions to the equation are 3 and -3.	or

The values 3 and -3 give a zero value in the denominator and must be excluded from the domain.

In set builder notation, the domain is delim{lbrace}{ x delim{ |} x!=3 and x!=-3}{rbrace}

In interval notation, the domain is (- infty,-3) union (-3,3) union (3,infty)

Find Vertical Asymptotes and Holes/Removable Discontinuities

A vertical asymptote occurs when there is a non-zero in the numerator and a zero in the denominator. To find the location of any vertical asymptotes, find the values where the denominator is zero but only after reducing the fraction so that we can guarantee that there is a non-zero in the numerator.

Factor the rational function and write it in lowest terms by canceling any common factors.	$f(x)= 1/{x^2-9}$
The rational function has no common factors and is in lowest terms	$f(x)= 1/{x^2-9}$
Find the values where the denominator is zero and the numerator is non-zero.
Solve the equation. This equation is a quadratic equation and can be solved by factoring, the square root method, completing the square or the quadratic formula.
Solve by factoring. Factor the expression on one side.
Use the zero product property and set each factor equal to zero.	or
Solve each equation.	or or or

The rational function has vertical asymptotes at x=-3 and x=3. That means that y approaches infty or - infty as x approaches the values of 3 and -3. Knowing this will guide us when sketching the graph.

Determine if the graph has any holes/removable discontinuities. In this example, since the rational function was in lowest terms there is no hole/removable discontinuity.

Find Horizontal Asymptotes and Oblique/Slant Asymptotes

Find the horizontal asymptotes using a set of rules based on the degree of the numerator and the degree of the denominator.

Horizontal Asymptote and Oblique/Slant Cases

Case 1: The degree of the numerator is less than the degree of the denominator.	The horizontal asymptote is y=0. There is no oblique/slant asymptote.
Case 2: The degree of the numerator is equal to the degree of the denominator.	The horizontal asymptote is y= the ratio of the leading coefficients. There is no oblique/slant asymptote.
Case 3: The degree of the numerator is one more than the degree of the denominator.	There is no horizontal asymptote. There is an oblique/slant asymptote. Find it using long division.
Case 4: The degree of the numerator is two or more than the degree of the denominator.	There is no horizontal asymptote and there is no oblique/slant asymptote.

For this example, the numerator is 1 and the degree of the numerator is 0. The denominator is x^2-9 and the degree of the numerator is 2. Since the degree of the numerator is less than the degree of the denominator (case 1) the horizontal asymptote is y=0. That means that y approaches 0 as x approaches infty or - infty . Knowing this will guide us when sketching the graph.

Find any Points that Cross the Horizontal or Oblique Asymptote

The horizontal and oblique asymptotes describe the graph as x approaches infty or - infty , usually called the end behavior. The graph will approach the asymptote but not reach the asymptote at the ends of the graph. However, the graph can cross the horizontal or oblique asymptote in the middle section of the graph. To find were the graph crosses the horizontal or oblique asymptote, set the value of the asymptote equal to the function.

For this example, set the horizontal asymptote (y=0) equal to the function $f(x)=1/{x^2-9}$ .

Horizontal Asymptote = Function

$0=1 /{x^2-9}$

$0(x^2-9)=1 /{x^2-9}(x^2-0)$

0=1

Since there is no solution to this equation, there are no points where the graph crosses the horizontal asymptote.

Find the Intercepts of the Rational Function

To find a y-intecept let x=0.

$f(0)=1/{0^2-9}$

f(0)=1/{-9}

f(0)=-{1/9}

The y-intercept is (0,-{1/9})

To find the x-intercept let y=0 or f(x)=0. Exclude solutions that were found to be undefined for the function when the domain was found.

$0=1 /{x^2-9}$

$0(x^2-9)=1 /{x^2-9}(x^2-0)$

0=1

Since this equation has no solution. There is no x-intercept.

Sketch the Graph

Draw the asymptotes as dashed lines. These will be used as guides to draw the graph. For this example, the vertical asymptotes were at x=-3 and x=3 and the horizontal asymptote was at y=0.

Plot the intercepts, holes or places where the graph crosses an asymptote. For this example, the only value is the y-intercept.

More points are needed to sketch the graph. Choose points in every section of the graph, using the vertical asymptotes to divide the sections up.

For example, choose x=-4 to represent the left section. The middle section needs a few extra points since it is surrounded by asymptotes. Choose x=-2 and x=2 to represent the middle section. Choose x=4 to represent the right section.

x	y
-4	$1/{(-4)^2-9}=1/{16-9}=1/7$
-2	$1/{(-2)^2-9}=1/{4-9}=1/{-5}=-{1/5}$
2	$1/{2^2-9}=1/{4-9}=1/{-5}=-{1/5}$
4	$1/{4^2-9}=1/{16-9}=1/7$

Plot the points on the graph.

Use the definition of the asymptotes to draw the graph near each point.

In the left section, consider the graph as x approaches - infty ( x right - infty ).

Since the graph has a horizontal asymptote at y=0, y right 0 as x right - infty . As you draw the graph to the left of -4, approach the horizontal asymptote.

In the left section, consider the graph as x approaches -3 from the left ( x right-3^- .)

Since the graph has a vertical asymptote at x=-3, y right infty or y right - infty as x right -3^- . Since the point in the left section lies above the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of -4 approach the vertical asymptote going towards positive infinity.

In the middle section, consider the graph as x approaches -3 from the right, ( x right-3^+ .)

Since the graph has a vertical asymptote at x=-3, y right infty or y right - infty as x right -3^+ . Since the point near -3 in the middle section lies below the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the left of -2, approach the vertical asymptote going towards negative infinity.

In the middle section, the three points can be connected to each other since there are no undefined values between them.

In the middle section, consider the graph as x approaches 3 from the left, ( x right 3^- .)

Since the graph has a vertical asymptote at x=3, y right infty or y right - infty as x right 3^- . Since the point near 3 in the middle section lies below the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of 2, approach the vertical asymptote going towards negative infinity.

In the right section, consider the graph as x approaches 3 from the right ( x right 3^+ .)

Since the graph has a vertical asymptote at x=3, y right infty or y right - infty as x right 3^+ . Since the point in the right section lies above the horizontal asymptote and it was determined that the graph does not cross the horizontal asymptote, . As you draw the graph to the right of 4 approach the vertical asymptote going towards positive infinity.

In the right section, consider the graph as x approaches infty ( x right infty ).

Since the graph has a horizontal asymptote at y=0, y right 0 as x right infty . As you draw the graph to the right of 4, approach the horizontal asymptote.

The sketch of the rational function $f(x)=1/{x^2-9}$ is below.

Here are some video examples that might help.

3.2 The Graphs of Functions

TI Calculator: Maximum, Minimum, and Zeros

February 26, 2018 admin

http://resources.chuh.org/staff/aklein/Alg2/TI_max%20min%20zeros.pdf

This is a link to screen shots and instructions for finding the minimum, maximum and zeros of a function using a TI graphing calculator.

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

Application of Quadratic Function: Maximize

November 17, 2017 admin

Example:

A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by the quadratic function h(t)=-4.9t^2+17.4t+1 . How long does it take the baseball to reach its maximum height? What is the maximum height obtained by the baseball?

Solution:

h(t)=-4.9t^2+14.7t+1 is defined to be a quadratic function. The graph of a quadratic function is a parabola. This quadratic function has a leading coefficient of -4.9 and since it is negative means that the parabola is opening down.

Find the vertex to find the maximum 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
Substitute the values of a and b into the formula
Simplify with a calculator