3.1 Analyzing Graphs Algebraically
Pre-Class:
- Take notes on the videos and readings (use the space below).
- Complete the 3.1 Pre-Class Quiz.
Preliminary Algebraic Analysis of the Function
Use Algebra and Limits to Identify All Basic Components in the Graph of $f(x).$
- Identify the Domain of $f(x)$: Commonly, $f(x)$is undefined for at any x value where:
- The denominator equals zero.
- There is an even root of a negative number.
- There is a logarithm of a negative number or log of zero.
- Identify x-intercepts and y-intercept of $f(x).$
- x-intercepts: Set $y = 0$ and solve for $x.$
- y-intercept: Set $x = 0$ and solve for $y.$
- Identify Vertical Asymptotes and Holes of $f(x).$
- Vertical asymptote at $a$ when $\mathop {\lim }\limits_{x \to a }f\left( x \right) = \frac{n}{0}$ if $n \ne 0.$
- Hole at $a$ when $\mathop {\lim }\limits_{x \to a }f\left( x \right) = \frac{0}{0}$ and then, after factoring and reducing, $\mathop {\lim }\limits_{x \to a }f\left( x \right) = \frac{n}{c}$ if $n \ne 0$ and $c$ is any real number.
- Identify Horizontal Asymptote of $f(x)$: calculate $\mathop {\lim }\limits_{x \to \infty } f\left( x \right).$
Notes
Use Algebra and Limits to Identify all the Basic Components of the Graph.
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$f(x)={{x}^{3}}+6{{x}^{2}}+9x$
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Domain:
$\left( -\infty ,\infty \right)$
Polynomial functions are always continuous.
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x int(s):
${{x}^{3}}+6{{x}^{2}}+9x=0$
$x\left( {{x}^{2}}+6x+9 \right)=0 $
$x{\left( x+3 \right)}{\left( x+3 \right)}=0 $
$x=0\,\,,\,\,x=-3$
$\left( 0,0 \right)\,\text{and}\,\left( -3,0 \right)$
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y int:
$f\left( 0 \right)={{\left( 0 \right)}^{3}}+6{{\left( 0 \right)}^{2}}+9\left( 0 \right)=0$
$\left( 0,0 \right)$
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Asymptotes:
There are no asymptotes. The function is a polynomial.
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$f(x)=\frac{3x+4}{2x-5}$
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Domain:
The domain will not contain x values where the denominator is equal to zero.
$ 2x-5\ne 0 $
$ x\ne \frac{5}{2}$
Domain: $\left( -\infty ,\frac{5}{2} \right)\cup \left( \frac{5}{2},\infty \right)$
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x int(s):
$0=\frac{3x+4}{2x-5}$
$\frac{0}{1}=\frac{3x+4}{2x-5}$
$0=3x+4$
$ x=-\frac{4}{3}$
$\left( -\frac{4}{3},0 \right)$
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y int:
$f\left( 0 \right)=\frac{3\left( 0 \right)+4}{2\left( 0 \right)-5}=-\frac{4}{5}$
$\left( 0,-\frac{4}{5} \right)$
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Asymptotes:
Vertical:
$\mathop {\lim }\limits_{x \to \frac{5}{2}}f\left( x \right)=\frac{11.5}{0}$
There is a vertical asymptote at $x=\frac{5}{2}$.
Horizontal:
$\mathop {\lim }\limits_{x \to \infty }\frac{3x+4}{2x+5}=\mathop {\lim }\limits_{x \to \infty}\frac{3x}{2x}= \mathop {\lim }\limits_{x \to \infty}\frac{3}{2}=\frac{3}{2}$
There is a horizontal asymptote at $y=\frac{3}{2}$.
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The annual first quarter change in revenue for Apple, Inc. is given in the table below.
Year % Revenue Growth 1998 -12.2 2000 27.1 2002 4.5 2004 29.4 2006 34.4 2008 42.7 2010 65.4 2012 58.9 2014 4.7 2016 -12.8 The regression model for this data is: $$f(x)=-0.005x^4+0.113x^3-0.889x^2+7.946x-5.346$$ where x is Years Since 1998.
Use Algebra and Limits to Identify All Basic Components in the Graph of $f(x)$:
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Identify the domain of $f(x)$:
[0,18]
This model shows 1st Quarter change in revenue from 1998 to 2016.
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Identify all x-intercepts in the domain:
(from calculator): $\left(0.727,0\right)$ and $\left(17.428,0\right)$
In general this means the percent change in revenue was 0% when x=0.727 and x=17.428 years after 1998. Specifically for this data, the percent change in revenue was negative in 1998 and positive in 2000 and then was positive in 2014 and negative in 2016.
Identify the y-intercept, if $x=0$ is in the domain:
$\left(0,-5.346\right)$
In 1998 percent change in revenue was approximately -5.346%, according to the model. The percent change in revenue was actually -12.2% in 1998.
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Identify all Asymptotes:
There are no vertical or horizontal asymptotes. This is a polynomial function.
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Using data from the Federal Reserve, the Dow S&P 500 annual percent return on investments for the years 2008-2014 can be modeled by the following equation: $$A(t) = - 1.64{t^4} + 20.85{t^3} - 86.05{t^2} + 127.87t - 36.24$$ where t is in years since 2008 and A(t) is in percent.
Source: http://pages.stern.nyu.edu/~adamodar/New_Home_Page/datafile/histretSP.html
Analyze and interpret the characteristics of the function.
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Identify the domain of $A(t).$
Domain: The interval $\left[ {0,6} \right]$
The variable t is in years since 2008. The model was developed for the years 2008-2014.
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Identify x-intercepts and y-intercepts of $A(t).$
Evaluate and interpret any x-intercepts of $A(t).$
$0 = - 1.64{t^4} + 20.85{t^3} - 86.05{t^2} + 127.87t - 36.24$
Using a graphing calculator, identify any zeros in the domain.
One x intercept identified in the domain.
$(0.3656,0)$
Sometime in the year 2008, returns reached 0%.
Evaluate and interpret the y intercept of $A(t)$.
$A\left( 0 \right) = - 1.64{\left( 0 \right)^4} + 20.85{\left( 0 \right)^3} - 86.05{\left( 0 \right)^2} + 127.87\left( 0 \right) - 36.24 = - 36.24$
$(0,-36.24)$
The annual percent return on investments in 2008 was -36.24.%
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Identify any vertical asymptotes and holes in $A(t).$
There are no vertical asymptotes or holes. The function is a polynomial.
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Identify any horizontal asymptotes in $A(t).$
There are no horizontal asymptotes. The function is a polynomial.
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