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Sketching the graph of a polynomial function

Sketch a graph of f ( x ) = −2 ( x + 3 ) 2 ( x 5 ) .

This graph has two x -intercepts. At x = −3 , the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x -intercept. At x = 5 , 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 f ( 0 ) .

f ( 0 ) = −2 ( 0 + 3 ) 2 ( 0 5 ) = −2 9 ( −5 ) = 90

The y -intercept is ( 0 , 90 ) .

Additionally, we can see the leading term, if this polynomial were multiplied out, would be 2 x 3 , 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] .

Showing the distribution for the leading term.

To sketch this, we consider that:

  • As x the function f ( x ) , so we know the graph starts in the second quadrant and is decreasing toward the x - axis.
  • Since f ( x ) = −2 ( x + 3 ) 2 ( x 5 ) is not equal to f ( x ) , the graph does not display symmetry.
  • At ( 3 , 0 ) , the graph bounces off of the x -axis, so the function must start increasing.

    At ( 0 , 90 ) , the graph crosses the y -axis at the y -intercept. See [link] .

Graph of the end behavior and intercepts, (-3, 0) and (0, 90), for the function f(x)=-2(x+3)^2(x-5).

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 ( 5 , 0 ) . See [link] .

Graph of the end behavior and intercepts, (-3, 0), (0, 90) and (5, 0), for the function f(x)=-2(x+3)^2(x-5).

As x the function f ( x ) −∞ , 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] .

Graph of f(x)=-2(x+3)^2(x-5).
The complete graph of the polynomial function f ( x ) = 2 ( x + 3 ) 2 ( x 5 )
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Sketch a graph of f ( x ) = 1 4 x ( x 1 ) 4 ( x + 3 ) 3 .

Graph of f(x)=(1/4)x(x-1)^4(x+3)^3.

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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 f whose graph is smooth and continuous. The Intermediate Value Theorem    states that for two numbers a and b in the domain of f , if a < b and f ( a ) f ( b ) , then the function f takes on every value between f ( a ) and f ( b ) . (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 f at x = a lies above the x - axis and another point at x = b lies below the x - axis, there must exist a third point between x = a and x = b where the graph crosses the x - axis. Call this point ( c ,   f ( c ) ) . This means that we are assured there is a solution c where f ( c ) = 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 x - axis. [link] shows that there is a zero between a and b .

Graph of an odd-degree polynomial function that shows a point f(a) that’s negative, f(b) that’s positive, and f(c) that’s 0.
Using the Intermediate Value Theorem to show there exists a zero.

Intermediate value theorem

Let f be a polynomial function. The Intermediate Value Theorem    states that if f ( a ) and f ( b ) have opposite signs, then there exists at least one value c between a and b for which f ( c ) = 0.

Questions & Answers

12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
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Carole
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
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
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
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salma
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salma
Commplementary angles
Idrissa Reply
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Sherica
im all ears I need to learn
Sherica
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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.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Practice Key Terms 4

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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