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

(c) Which function belongs to both families?

Short answer

(a) The equation for the family of linear functions with slope 2 is \(y = 2x + b\). The sketch is shown below:

(b) The equation is \(y = mx + \left( {1 - 2m} \right)\). The sketch is shown below:

(c) The function \(y = 2x - 3\) is the only one which belongs to the both families.

Step-by-step solution

The slope-intercept form

Recall that the equation \(y = mx + b\) is the slope-intercept equation form of a line, where\(m\)is the slope and\(b\)is the intercept.

It is given that the slope is 2. So, the slope is\(m = 2\).

The equation becomes as shown below:

\(y = 2x + b\)

Here,\(b\)represents the\(y\)-intercept.

Take\(b = - 1\), and substitute in the equation\(y = 2x + b\)as shown below:

\(y = 2x - 1\)

Take\(b = 0\), and substitute in the equation\(y = 2x + b\)as shown below:

\(y = 2x\)

Take\(b = 3\), and substitute in the equation\(y = 2x + b\)as shown below:

\(y = 2x + 3\)

The graph of the members \(y = 2x - 1\), \(y = 2x\), and \(y = 2x + 3\) is shown below:

The point-slope form

It is given that \(f\left( 2 \right) = 1\) by comparing with the function \(f\left( x \right) = y\). It can be represented as in point form \(\left( {2,1} \right)\).

Recall that the point slope equation form is\(y - {y_1} = m\left( {x - {x_1}} \right)\), where\(\left( {{x_1},{y_1}} \right)\)is the point and\(m\)is the slope.

The equation becomes as shown below:

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

The equation\(y = mx + \left( {1 - 2m} \right)\)is the slope-intercept form.

For \(m = 0\), the equation is shown below:

\(y = 1\)

For \(m = 1\), the equation is shown below:

\(y = x - 1\)

For \(m = - 1\), the equation is shown below:

\(y = - x + 3\)

The graph of the members\(y = 1\),\(y = x - 1\), and\(y = - x + 3\)is shown below:

The function belongs to the family

For \(m = 2\), the function belongs to both the families.

From part (b), the equation is\(y = mx + \left( {1 - 2m} \right)\).

So, for \(m = 2\), the equation is shown below:

\(y = 2x - 3\)

Thus, it is the only one which belongs to the both families.