5.3 Graphs of polynomial functions  (Page 8/13)

 Page 8 / 13

Access the following online resource for additional instruction and practice with graphing polynomial functions.

Key concepts

• Polynomial functions of degree 2 or more are smooth, continuous functions. See [link] .
• To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. See [link] , [link] , and [link] .
• Another way to find the $\text{\hspace{0.17em}}x\text{-}$ intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the $\text{\hspace{0.17em}}x\text{-}$ axis. See [link] .
• The multiplicity of a zero determines how the graph behaves at the $\text{\hspace{0.17em}}x\text{-}$ intercepts. See [link] .
• The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
• The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
• The end behavior of a polynomial function depends on the leading term.
• The graph of a polynomial function changes direction at its turning points.
• A polynomial function of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ has at most $\text{\hspace{0.17em}}n-1\text{\hspace{0.17em}}$ turning points. See [link] .
• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most $\text{\hspace{0.17em}}n-1\text{\hspace{0.17em}}$ turning points. See [link] and [link] .
• Graphing a polynomial function helps to estimate local and global extremas. See [link] .
• The Intermediate Value Theorem tells us that if have opposite signs, then there exists at least one value $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ for which $\text{\hspace{0.17em}}f\left(c\right)=0.\text{\hspace{0.17em}}$ See [link] .

Verbal

What is the difference between an $\text{\hspace{0.17em}}x\text{-}$ intercept and a zero of a polynomial function $\text{\hspace{0.17em}}f?\text{\hspace{0.17em}}$

The $\text{\hspace{0.17em}}x\text{-}$ intercept is where the graph of the function crosses the $\text{\hspace{0.17em}}x\text{-}$ axis, and the zero of the function is the input value for which $\text{\hspace{0.17em}}f\left(x\right)=0.$

If a polynomial function of degree $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct zeros, what do you know about the graph of the function?

Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.

If we evaluate the function at $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and at $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and the sign of the function value changes, then we know a zero exists between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.$

Explain how the factored form of the polynomial helps us in graphing it.

If the graph of a polynomial just touches the x -axis and then changes direction, what can we conclude about the factored form of the polynomial?

There will be a factor raised to an even power.

Algebraic

For the following exercises, find the $\text{\hspace{0.17em}}x\text{-}$ or t -intercepts of the polynomial functions.

$\text{\hspace{0.17em}}C\left(t\right)=2\left(t-4\right)\left(t+1\right)\left(t-6\right)$

$\text{\hspace{0.17em}}C\left(t\right)=3\left(t+2\right)\left(t-3\right)\left(t+5\right)$

$\left(-2,0\right),\left(3,0\right),\left(-5,0\right)$

$\text{\hspace{0.17em}}C\left(t\right)=4t{\left(t-2\right)}^{2}\left(t+1\right)$

$\text{\hspace{0.17em}}C\left(t\right)=2t\left(t-3\right){\left(t+1\right)}^{2}$

$\text{\hspace{0.17em}}\left(3,0\right),\left(-1,0\right),\left(0,0\right)$

$\text{\hspace{0.17em}}C\left(t\right)=2{t}^{4}-8{t}^{3}+6{t}^{2}$

$\text{\hspace{0.17em}}C\left(t\right)=4{t}^{4}+12{t}^{3}-40{t}^{2}$

$\text{\hspace{0.17em}}f\left(x\right)={x}^{4}-{x}^{2}$

$\text{\hspace{0.17em}}f\left(x\right)={x}^{3}+{x}^{2}-20x$

$f\left(x\right)={x}^{3}+6{x}^{2}-7x$

$f\left(x\right)={x}^{3}+{x}^{2}-4x-4$

$f\left(x\right)={x}^{3}+2{x}^{2}-9x-18$

$f\left(x\right)=2{x}^{3}-{x}^{2}-8x+4$

$\left(-2,0\right),\text{\hspace{0.17em}}\left(2,0\right),\text{\hspace{0.17em}}\left(\frac{1}{2},0\right)$

$f\left(x\right)={x}^{6}-7{x}^{3}-8$

$f\left(x\right)=2{x}^{4}+6{x}^{2}-8$

$f\left(x\right)={x}^{3}-3{x}^{2}-x+3$

$f\left(x\right)={x}^{6}-2{x}^{4}-3{x}^{2}$

$\left(0,0\right),\text{\hspace{0.17em}}\left(\sqrt{3},0\right),\text{\hspace{0.17em}}\left(-\sqrt{3},0\right)$

$f\left(x\right)={x}^{6}-3{x}^{4}-4{x}^{2}$

$f\left(x\right)={x}^{5}-5{x}^{3}+4x$

For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.

$f\left(x\right)={x}^{3}-9x,\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}x=-4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-2.$

$f\left(x\right)={x}^{3}-9x,\text{\hspace{0.17em}}$ between $\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=4.$

$f\left(2\right)=–10\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(4\right)=28.$ Sign change confirms.

Questions & Answers

if tan alpha + beta is equal to sin x + Y then prove that X square + Y square - 2 I got hyperbole 2 Beta + 1 is equal to zero
Rahul Reply
sin^4+sin^2=1, prove that tan^2-tan^4+1=0
SAYANTANI Reply
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Kuz
A = P(1 + r/n) ^rt
Dale
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Kavita
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kurash
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Kavita
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Masum
What is the value of log-1
Masum
the value of log1=0
Kavita
Log(-1)
Masum
What is the value of i^i
Masum
log -1 is 1.36
kurash
No
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Kavita
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Masum
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Kavita
tan20°×tan30°×tan45°×tan50°×tan60°×tan70°
Joju Reply
jaldi batao
Joju
Find the value of x between 0degree and 360 degree which satisfy the equation 3sinx =tanx
musah Reply
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tae Reply
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Sanjana Reply
1×10^0
Akugry
Evalute exponential functions
Sujata Reply
30
Shani
The sides of a triangle are three consecutive natural number numbers and it's largest angle is twice the smallest one. determine the sides of a triangle
Jaya Reply
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Inkoom
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Neese
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Gaurav
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Akugry
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Akugry
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Stormzy
x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
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x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
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ranges
EDWIN
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Oliver
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Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
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