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Using the intermediate value theorem

Show that the function f ( x ) = x 3 5 x 2 + 3 x + 6 has at least two real zeros between x = 1 and x = 4.

As a start, evaluate f ( x ) at the integer values x = 1 , 2 , 3 , and 4. See [link] .

x 1 2 3 4
f ( x ) 5 0 –3 2

We see that one zero occurs at x = 2. Also, since f ( 3 ) is negative and f ( 4 ) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between x = 1 and x = 4.

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Show that the function f ( x ) = 7 x 5 9 x 4 x 2 has at least one real zero between x = 1 and x = 2.

Because f is a polynomial function and since f ( 1 ) is negative and f ( 2 ) is positive, there is at least one real zero between x = 1 and x = 2.

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Writing formulas for polynomial functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function    written in factored form will have an x -intercept where each factor is equal to zero, we can form a function that will pass through a set of x -intercepts by introducing a corresponding set of factors.

Factored form of polynomials

If a polynomial of lowest degree p has horizontal intercepts at x = x 1 , x 2 , , x n , then the polynomial can be written in the factored form: f ( x ) = a ( x x 1 ) p 1 ( x x 2 ) p 2 ( x x n ) p n where the powers p i on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x -intercept.

Given a graph of a polynomial function, write a formula for the function.

  1. Identify the x -intercepts of the graph to find the factors of the polynomial.
  2. Examine the behavior of the graph at the x -intercepts to determine the multiplicity of each factor.
  3. Find the polynomial of least degree containing all the factors found in the previous step.
  4. Use any other point on the graph (the y -intercept may be easiest) to determine the stretch factor.

Writing a formula for a polynomial function from the graph

Write a formula for the polynomial function shown in [link] .

Graph of a positive even-degree polynomial with zeros at x=-3, 2, 5 and y=-2.

This graph has three x -intercepts: x = −3 , 2 , and 5. The y -intercept is located at ( 0 , 2 ) . At x = −3 and x = 5 , the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2 , the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

f ( x ) = a ( x + 3 ) ( x 2 ) 2 ( x 5 )

To determine the stretch factor, we utilize another point on the graph. We will use the y - intercept ( 0 , 2 ) , to solve for a .

f ( 0 ) = a ( 0 + 3 ) ( 0 2 ) 2 ( 0 5 ) −2 = a ( 0 + 3 ) ( 0 2 ) 2 ( 0 5 ) −2 = 60 a a = 1 30

The graphed polynomial appears to represent the function f ( x ) = 1 30 ( x + 3 ) ( x 2 ) 2 ( x 5 ) .

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Given the graph shown in [link] , write a formula for the function shown.

Graph of a negative even-degree polynomial with zeros at x=-1, 2, 4 and y=-4.

f ( x ) = 1 8 ( x 2 ) 3 ( x + 1 ) 2 ( x 4 )

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Using local and global extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Questions & Answers

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
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
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
kkk nice
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
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
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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