# 5.3 Graphs of polynomial functions  (Page 5/13)

 Page 5 / 13

## Sketching the graph of a polynomial function

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x-5\right).$

This graph has two x -intercepts. At $\text{\hspace{0.17em}}x=-3,\text{\hspace{0.17em}}$ the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x -intercept. At $\text{\hspace{0.17em}}x=5,\text{\hspace{0.17em}}$ the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y -intercept is found by evaluating $\text{\hspace{0.17em}}f\left(0\right).$

$\begin{array}{ccc}\hfill f\left(0\right)& =& -2{\left(0+3\right)}^{2}\left(0-5\right)\hfill \\ & =& -2\cdot 9\cdot \left(-5\right)\hfill \\ & =& 90\hfill \end{array}$

The y -intercept is $\text{\hspace{0.17em}}\left(0,90\right).$

Additionally, we can see the leading term, if this polynomial were multiplied out, would be $\text{\hspace{0.17em}}-2{x}^{3},\text{\hspace{0.17em}}$ so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See [link] .

To sketch this, we consider that:

• As $\text{\hspace{0.17em}}x\to -\infty \text{\hspace{0.17em}}$ the function $\text{\hspace{0.17em}}f\left(x\right)\to \infty ,\text{\hspace{0.17em}}$ so we know the graph starts in the second quadrant and is decreasing toward the $\text{\hspace{0.17em}}x\text{-}$ axis.
• Since $\text{\hspace{0.17em}}f\left(-x\right)=-2{\left(-x+3\right)}^{2}\left(-x–5\right)\text{\hspace{0.17em}}$ is not equal to $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ the graph does not display symmetry.
• At $\text{\hspace{0.17em}}\left(-3,0\right),\text{\hspace{0.17em}}$ the graph bounces off of the x -axis, so the function must start increasing.

At $\text{\hspace{0.17em}}\left(0,90\right),\text{\hspace{0.17em}}$ the graph crosses the y -axis at the y -intercept. See [link] .

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at $\text{\hspace{0.17em}}\left(5,0\right).\text{\hspace{0.17em}}$ See [link] .

As $\text{\hspace{0.17em}}x\to \infty \text{\hspace{0.17em}}$ the function $\text{\hspace{0.17em}}f\left(x\right)\to \mathrm{-\infty },\text{\hspace{0.17em}}$ so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in [link] , and verify that the resulting graph looks like our sketch in [link] .

Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{4}x{\left(x-1\right)}^{4}{\left(x+3\right)}^{3}.$

## Using the intermediate value theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x -axis, we can confirm that there is a zero between them. Consider a polynomial function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ whose graph is smooth and continuous. The Intermediate Value Theorem    states that for two numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ in the domain of $\text{\hspace{0.17em}}f,$ if $\text{\hspace{0.17em}}a and $f\left(a\right)\ne f\left(b\right),$ then the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ takes on every value between $\text{\hspace{0.17em}}f\left(a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(b\right).\text{\hspace{0.17em}}$ (While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ lies above the $\text{\hspace{0.17em}}x\text{-}$ axis and another point at $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ lies below the $\text{\hspace{0.17em}}x\text{-}$ axis, there must exist a third point between $\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=b\text{\hspace{0.17em}}$ where the graph crosses the $\text{\hspace{0.17em}}x\text{-}$ axis. Call this point This means that we are assured there is a solution $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ where $f\left(c\right)=0.$

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the $\text{\hspace{0.17em}}x\text{-}$ axis. [link] shows that there is a zero between $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$

## Intermediate value theorem

Let $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ be a polynomial function. The Intermediate Value Theorem    states that if $\text{\hspace{0.17em}}f\left(a\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(b\right)\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}f\left(c\right)=0.$

#### Questions & Answers

12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8