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2 years ago

15

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?

2 answers:

babunello [35]2 years ago

6 0

Answer:

x + 6

Step-by-step explanation:

Edit:

A moderator mentioned my answer had a mistake: turned out the question posted was incomplete. The complete question should read:

"

What must be a factor of the polynomial function f(x) graphed on the coordinate plane below?

x – 6

x – 3

x + 1

x + 6

"

As shown in the same graph, f(x) cross the x-axis at x = -6, -3 and 0.

When a polynomial crosses the x-axis at x = a, (x - a) must be a factor.

So (x + 6), (x + 3) and x are factors of f(x).

Of the options, the correct answer is x + 6.

The complete question with verified answer is here: brainly.com/question/3875529

Harman [31]2 years ago

4 0

First let's check for zeros

Mark the x intercepts

-6,-3,0

The factors are

x+6
x+3
x

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It is estimated that there are 828 motor vehicles in the US it is estimated that there are 828 Motor vehicles in the US for ever

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Step-by-step explanation:

Given that :

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Then,

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3 years ago

There are eight boys and five girls in Chandler's group. Which fraction represents the part that is boys? A:5/8 B:8/13 C:5/13 D:

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3 years ago

Which point on the graph of g(x)=(1/5)^x? HELPP

Cloud [144]

Answer:

(-1,5) and $(3, \frac{1}{125})$ are points on the graph

Step-by-step explanation:

Given

$g(x) = \frac{1}{5}^x$

Required

Determine which point in on the graph

To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))

A. (-1,5)

x = -1

Substitute -1 for x in $g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-1}$

Convert to index form

$g(x) = 1/(\frac{1}{5})$

Change / to *

$g(x) = 1*(\frac{5}{1})$

$g(x) = 5$

This satisfies (-1,5)

<em>Hence, (-1,5) is on the graph</em>

<em></em>

B. (1,0)

x = 1

Substitute 1 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^1$

$g(x) = \frac{1}{5}$

<em>(1,0) is not on the graph because g(x) is not equal to 0</em>

C. $(3, \frac{1}{125})$

x = 3

Substitute 3 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^3$

Apply law of indices

$g(x) = \frac{1}{5} * \frac{1}{5} * \frac{1}{5}$

$g(x) = \frac{1}{125}$

This satisfies $(3, \frac{1}{125})$

<em>Hence, </em> $(3, \frac{1}{125})$ <em> is on the graph</em>

<em></em>

D. $(-2, \frac{1}{25})$

x = -2

Substitute -2 for x

$g(x) = \frac{1}{5}^x$

$g(x) = \frac{1}{5}^{-2}$

Convert to index form

$g(x) = 1/(\frac{1}{5}^2)$

$g(x) = 1/(\frac{1}{5}*\frac{1}{5})$

$g(x) = 1/(\frac{1}{25})$

Change / to *

$g(x) = 1*(\frac{25}{1})$

$g(x) = 25$

This does not satisfy $(-2, \frac{1}{25})$

<em>Hence, </em> $(-2, \frac{1}{25})$ <em> is not on the graph</em>

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4 years ago