5.3 Graphs of polynomial functions  (Page 8/13)

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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.

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
SO THE ANSWER IS X=-8
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
1KI POWER 1/3 PLEASE SOLUTIONS
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1