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Mathematics, 22.01.2020 00:31 hendersontianna05

(a) find an equation for the family of linear functions with slope 2 and sketch several members of the family.
(b) find an equation for the family of linear functions such that (2) = 1 and sketch several members ofthe family.
(c) which function belongs to both families?

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Answer from: jpsaad00

$\textbf{(\text{a})}$ f(x) = 2x+n

$\textbf{(\text{b})}\\$ f(x) = mx + 1 -2m = m(x-2) + 1

$\textbf{(\text{c})}$ f(x) = 2x - 3

Step-by-step explanation:

First, we know that family of functions represents a set of functions whose equations have a similar form. In our case, a family of linear functions can be represented as

$\{ ax + b | a, b \in \mathbb{R} \}$ .

Now, we can take an arbitrary member of that family, a function

f(x) = mx + n

for some real constants and .

$\textbf{(\text{a})}$

In this part of the problem, we know that $m = 2\\$ , so we consider

f(x) = 2x +n .

To graph several members of the family, you can plug in any real number in the equation above instead of , since $\forall n \in \mathbb{R}$ satisfy the equation.

For n = 0 , we have f(x) = 2x.

For n = 15 , we have f(x) = 2x + 15 .

For n = -15 , we have f(x) = 2x - 15.

The graphs for the values n = 0, n = 15 and n = -15 are presented on the first graph below.

$\textbf{(\text{b})}$

We need to find the member of the family of linear functions such that

f(2) = 1 .

Substituting for in f(x) = mx + n gives

f(2) = 2m+n .

Now, since we have that f(2) = 1 , we can equate with 2m +n and express one of them in terms of the other.

$2m + n = 1 \implies n = 1 - 2m$

Substituting 1-2m for in f(x) = mx + n gives the equation

f(x) = mx + 1 -2m = m(x-2) + 1

which represents the wanted family. To sketch several member, we can choose any real value for , since $\forall m \in \mathbb{R}$ satisfy the equation.

For m = 0 , we have f(x) = 1 .

For m = 5 , we have f(x) = 5x - 9.

The graph is presented below.

$\textbf{(\text{c})}$

A function belongs to both families if it satisfies both conditions; Its slope must be equal to and f(2) = 1 .

Let's consider a function

f(x) = mx + n

for some real constants and .

The objective is to find the numeric value of the constants and . Since the slope must be equal to , we obtain that m = 2 and

f(x) = 2x + n .

To find the numeric value of , we use the fact that f(2) = 1 .

Substituting for gives

$f(2) = 2 \cdot 2 + n = 4 + n$ .

On the other hand, since f(2) = 1 , we obtain that

$4 + n = f(2) = 1 \implies 4 + n = 1 \implies n = 1 - 4 \implies n = -3$

Therefore, a function that belongs to both families is

f(x) = 2x - 3 .

(a) find an equation for the family of linear functions with slope 2 and sketch several members of t

(a) find an equation for the family of linear functions with slope 2 and sketch several members of t

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Answer from: Quest

irrational, cannot

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i believe it is d.) buddhism and e.) hinduism

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