Question

( 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\({\bf{f}}\left( {\bf{2}} \right){\bf{ = 1}}\)and sketch several members of the family.

(c) Which function belongs to both families?

Short answer

1. The equation for the family of the linear function with slope 2 is
2. The equation for the family of linear functions which passes through the point \(\left( {2,1} \right)\) is .
3. The function that belongs to both the families is.

Step-by-step solution

(a) Determine an equation for the family of linear functions with slope

The objective is to find an equation for the family of a linear function with slope 2.

Consider the linear function of the form,\(y = mx + b\).

Here,\(m\) is the slope and \(b\)is the y-intercept.

Substitute the slope\(m = 2\)in the above linear function.

Therefore, the equation for the family of the linear function with slope 2

is\(y = 2x + b\).

Find the members of the family of the linear function by substituting any real numbers in place of b.

Substitute\(b = 0\)in\(y = 2x + b\)\(\)

\(y = 2x + 0\)

\(y = 2x\)

Substitute\(b = 2\)in\(y = 2x + b\)

\(y = 2x + 2\)

Substitute\(b = - 3\)in\(y = 2x + b\)

\(y = 2x - 3\)

\(\)

Step 2:The graph is given below

Now sketch the graph of the members of the family of\(y = 2x + b\) as follows:

(b) Determine and equation for the family of linear functions such that \({\bf{f}}\left( {\bf{2}} \right){\bf{ = 1}}\)

The objective is to find an equation for the family of linear functions such that\(f\left( 2 \right) = 1\)

The linear function passes through the point\(\left( {2,1} \right)\), since\(f\left( 2 \right) = 1\)

Substitute the point in the point-slope form,

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

\(y - 1 = m\left( {x - 2} \right)\)

\(y = mx - 2m + 1\)

Therefore, the equation for the family of linear functions which passes through the point \(\left( {2,1} \right)\) is,.

Find the members of the family of the linear function by substituting any real numbers in place of m.

Substitute\(m = - 1\),

\(\begin{array}{c}y = mx - 2m + 1\\ = \left( { - 1} \right)x - 2\left( { - 1} \right) + 1\\ = - x + 3\end{array}\)

Substitute\(m = 0\)in\(y = mx - 2m + 1\)

\(\begin{array}{c}y = mx - 2m + 1\\ = \left( 0 \right)x - 2\left( 0 \right) + 1\\ = 1\end{array}\)

Substitute\(m = 1\)in\(y = mx - 2m + 1\)

\(\begin{array}{c}y = mx - 2m + 1\\ = \left( 1 \right)x - 2\left( 1 \right) + 1\\ = x - 1\end{array}\)

The graph is given below

Now sketch the graph of the members of the family of\(y = mx - 2m + 1\)as follow;

(c) Determine which function belongs to both families

The objective is to find the function which belongs to both the families in part (a) and part (b).

To belong to both families the slope of the function should be 2 and it should pass through the point\(\left( {2,1} \right)\)

Substitute\(m = 2\)and the point\(\left( {2,1} \right)\)in the slope-intercept form,

\(\begin{array}{c}y - y' = m\left( {x - {x_1}} \right)\\y - 1 = 2\left( {x - 2} \right)\\y - 1 = 2x - 4\\y = 2x - 3\end{array}\)

Therefore, the function that belongs to both families is.

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