Question

What do all members of the family of linear functions \(f\left( x \right) = {\bf{1}} + m\left( {x + {\bf{3}}} \right)\) have in common? Sketch several members of the family.

Short answer

All the members of the family of linear functions pass through the point \(\left( { - 3,1} \right)\). The sketch of several members of the family is shown below:

Step-by-step solution

The point-slope form

It is given that \(f\left( x \right) = 1 + m\left( {x + 3} \right)\), or \(y = 1 + m\left( {x + 3} \right)\).

Rearrange the equation\(y = 1 + m\left( {x + 3} \right)\)as shown below:

\(y - 1 = m\left( {x + 3} \right)\)

Recall that thepoint 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.

Compare the equation\(y - 1 = m\left( {x + 3} \right)\)with the general equation\(y - {y_1} = m\left( {x - {x_1}} \right)\). So,\(\left( {{x_1},{y_1}} \right) \equiv \left( { - 3,1} \right)\).

Thus, all the members of the family of linear functions cross the point \(\left( { - 3,1} \right)\).

Sketch the graph

The equation \(y = m\left( {x + 3} \right) + 1\) is the slope-intercept form.

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

\(y = 1\)

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

\(y = x + 4\)

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

\(y = - x - 2\)

For \(m = \frac{1}{2}\), the equation is shown below:

\(\begin{aligned}y &= \frac{1}{2}\left( {x + 3} \right) + 1\\ &= \frac{x}{2} + \frac{5}{2}\\ &= \frac{1}{2}\left( {x + 5} \right)\end{aligned}\)

The graph of the members \(y = 1\), \(y = x + 4\), \(y = - x - 2\) and \(y = \frac{1}{2}\left( {x + 5} \right)\) is shown below:

Thus, the sketch is shown above.