Category Archives: Trigsted Chapter 4

Analyze a Quadratic Function in Vertex Form

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
f(0)=-{1/4}(1)+5
Multiply.
f(0)=-{1/4}(1)+5
f(0)=-{1/4}+5
Get a common denominator and combine fractions
f(0)=-{1/4}+5
f(0)=-{1/4}+20/4
f(0)=19/4

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
20=(x-1)^2
Get rid of the square by square rooting both sides.
20=(x-1)^2
sqrt{20}=sqrt{(x-1)^2}
pm sqrt{20}=x-1
Simplify the radical by finding a perfect square factor of 20.
pm sqrt{20}=x-1
pm sqrt{4*5}=x-1
pm 2 sqrt{5}=x-1
Isolate x by adding 1 to both sides.  Then simplify.
pm 2 sqrt{5}=x-1
1 pm 2 sqrt{5}=x-1+1
1  pm   2 sqrt{5}=x
Rewrite as two separate answers.
1  pm   2 sqrt{5}=x
1 + 2 sqrt{5}=x or 1 - 2 sqrt{5}=x

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.

Application: Quadratic Minimimization

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 4x 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 6y 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
4x+6y=28
Solve for one of the variables.  In this example, solve for x by first subtracting 6y from both sides. Simplify.
4x+6y=28
4x+6y-6y=28-6y
4x=28-6y
Continue to solve for x by dividing both sides by 4 and simplifying.
4x=28-6y
{4x}/4={28-6y}/4
x=28/4-{6y}/4
x=7-{3/2}y

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

Area Expression
A=2y^2+x^2
Substitute for x.
A=2y^2+x^2
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
y=-b/{2a}
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
a={17/4}; b=-21; c=49
Substitute the values of a and b into the formula
y=-b/{2a}
y={-(-21)}/{2({17/4})}
y={21}/{34/4}
y={21}/{17/2}
y={21}*{2/17}
y=42/17
Simplify with a calculator
y=42/17
y=2.470588235

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.

Rational Function: Sketch the graph of the rational function

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.
x^2-9=0
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. x^2-9=0
Solve by the square root method. Start by isolating the square. x^2-9=0
x^2-9+9=0+9
x^2=9
Cancel out the square by square rooting both sides. x^2=9
sqrt{x^2}=sqrt{9}
x=pm 3
The solutions to the equation are 3 and -3. x=-3 or x=3

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}
f(x)= 1/{(x+3)(x-3)}
The rational function has no common factors and is in lowest terms
f(x)= 1/{x^2-9}
f(x)= 1/{(x+3)(x-3)}
Find the values where the denominator is zero and the numerator is non-zero.
x^2-9=0
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. x^2-9=0
Solve by factoring.  Factor the expression on one side. x^2-9=0
(x+3)(x-3)=0
Use the zero product property and set each factor equal to zero. (x+3)(x-3)=0
x+3=0 or x-3=0
Solve each equation. x+3=0 or x-3=0
x+3-3=0-3 or x-3+3=0+3
x=-3 or x=3

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.
n<m
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.
n=m
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.
n=m+1
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, y right infty.  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, y right - infty.  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, y right - infty.  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, y right infty.  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.

Application of Quadratic Function: Maximize

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
t=-b/{2a}
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
h(t)=-4.9t^2+14.7t+1
a=-4.9; b=14.7; c=1
Substitute the values of a and b into the formula
t=-b/{2a}
t={-(14.7)}/{2(-4.9)}
Simplify with a calculator
t={-(14.7)}/{2(-4.9)}
t=1.5

The ball reaches the maximum height 1.5 seconds after the ball was hit.

The maximum height can be found by substituting 1.5 seconds for time in the height function.

The height function
h(t)=-4.9t^2+14.7t+1
Substitute t=1.5 in the height function
h(t)=-4.9t^2+14.7t+1
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
Simplify using a calculator
h(1.5)=-4.9(1.5)^2+14.7(1.5)+1
h(1.5)=12.025

The maximum height of the ball is 12.025 meters.

Here is a video example maximizing the position function.

 

Finding Domain: Rational Function

Example:  Classify the function as a polynomial function, rational function, or root function, and then find the domain.  Write the domain interval notation and set builder notation.

h(x) = {x^2+8}/{x^2+x-20}

Solution:

Classify the Function

Formal Definition
Practical Way of Identifying
Polynomial Function

A polynomial function is a function of the form

f(x)=a_n x^n+a_{n-1} x^{n-1}+...+a_1 x+a_0

where n is a non-negative integer {0, 1, 2, 3, 4, …} and the coefficients a_n , a_{n-1,...,a_1,a_0 are from the real numbers.

Look for the variables to be in the numerator.  (If there is no fraction at all, they are in the numerator.)  The variable should not be inside a radical or absolute value.  The powers or exponents on the variables should be whole numbers.  Whole numbers come from this list {0, 1, 2, 3, 4, …}
Rational Function

A rational function is a function of the form

f(x)={p(x)}/{q(x)}

where p(x) and q(x) are polynomial functions and q(x) is not equal to zero.

The numerator and denominator are polynomials.  Most functions with variables in the denominator are considered Rational Functions but there are exceptions.
 Root Function (even index)

A root function is a function of the form

where n is an even positive integer greater than or equal to 2.

 The variable is inside or underneath a radical.  The index of the radical is an even number.  {2, 4, 6, 8, …}  The square root is an even index although the index is not written.
 Root Function (odd index)

A root function is a function of the form

where n is an odd positive integer greater than or equal to 2.

The variable is inside or underneath a radical.  The index of the radical is an odd number.  {3, 5, 7, 9, …}  The cube root is an odd index.

Since the function h(x) = {x^2+8}/{x^2+x-20} has a variable in the denominator and the numerator and denominator are polynomial functions this function is a rational function.

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.
x^2+x-20=0
Solve the equation. This equation is a quadratic equation and can be solved by factoring, completing the square or the quadratic formula. x^2+x-20=0
Solve by factoring.  Factor the expression on one side. x^2+x-20=0
(x+5)(x-4)=0
Use the zero product property and set each factor equal to zero. (x+5)(x-4)=0
x+5=0 or x-4=0
Solve each equation. x+5=0 or x-4=0
x+5-5=0-5 or x-4+4=0+4
x=-5 or x=4

The values -5 and 4 give a zero value in the denominator, make the function undefined and must be excluded from the domain.

In set builder notation, the domain is delim{lbrace}{ x delim{ |} x!=-5 and x!=4}{rbrace}

In interval notation, the domain is (- infty,-5) union (-5,4) union (4,infty)

 

Quadratic Function: General Form

Example: Rewrite the given quadratic function in standard form by completing the square.  Then 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)=-x^2-2x+8

Solution:

Rewrite function in standard form.

f(x)=-x^2-2x+8

f(x)=-(x^2+2x)+8

f(x)=-(x^2+2x+1)+8+1

f(x)=-(x+1)^2+9

Here is a youTube video that demonstrations the process.

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)=-(x+1)^2+9

The vertex of the quadratic function is (-1,9).

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.

f(0)=-(0+1)^2+9

f(0)=-(1)^2+9

f(0)=-1+9

f(0)=8

The y-intercept is (0,8).

Find the x-intercept.

To find an x-intercept let y=0 or f(x)=0.

0=-(x+1)^2+9

0-9=-(x+1)^2+9-9

-9=-(x+1)^2

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

9=(x+1)^2

sqrt{9}=sqrt{(x+1)^2}

pm 3=x+1

-1 pm 3=x+1-1

-1 pm 3=x

-1 + 3=x or -1 - 3=x

2=x or -4=x

Find the axis of symmetry.

The axis of symmetry is a vertical line that passes through the vertex.  Since the vertex is (-1,9) 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, 9 {}{]}.

Here a few video examples of analyzing the graph in general form.