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2 years ago

Consider the function f(x) = 1/3(6)x. What is the value of the growth factor of the function?

Mathematics

2 answers:

Ganezh [65]2 years ago

8 0

The answer is c to this question

Ostrovityanka [42]2 years ago

8 0

Answer:

growth factor is 6

Step-by-step explanation:

$f(x) =\frac{1}{3} (6)^x$

For any equation f(x)=ab^x

'a' is the initial value

b is the growth or decay factor .

When the value of b is greater than 1 then its growth factor

when the value of b is between 0 and 1 then it is decay factor

x is the number of years

In our given function

$f(x) =\frac{1}{3} (6)^x$

the value of b is 6

so growth factor is 6

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What is the answer please.... need help

mina [271]

Answer:

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

$f\left(x\right)=x^3-6x^2+3x+10$

As the highest power of the x-variable is 3 with the leading coefficients of 1.

So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

$f\left(x\right)=x^3-6x^2+3x+10$

$0=x^3-6x^2+3x+10$ ∵ $f(x)=0$

$Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

$\left(x+1\right)\left(x-2\right)\left(x-5\right)=0$

Using the zero factor principle

if $ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0$

$x=-1,\:x=2,\:x=5$

Therefore, the zeros of the function are:

$x=-1,\:x=2,\:x=5$

$f\left(x\right)=x^3-6x^2+3x+10$ is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.

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