Question

Draw the graph of y= mod x+ 1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\red{\rm :\longmapsto\:y = |x| + 1}$

Let first define the function,

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{x \: \: when \: x \geqslant 0} \\ &\sf{ - x \: when \: x < 0} \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| + 1= \begin{cases} &\sf{x + 1 \: \: when \: x \geqslant 0} \\ &\sf{ - x + 1 \: when \: x < 0} \end{cases}\end{gathered}\end{gathered}$

So, it means,

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: y = |x| + 1= \begin{cases} &\sf{x + 1 \: \: when \: x \geqslant 0} \\ &\sf{ - x + 1 \: when \: x < 0} \end{cases}\end{gathered}\end{gathered}}$

Now,

Consider

$\rm :\longmapsto\:y = x + 1 \: \: when \: x \: \geqslant 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:y = 0 + 1$

$\rm :\longmapsto\:y =1$

Substituting 'x = 1' in the given equation, we get

$\rm :\longmapsto\:y = 1 + 1$

$\rm :\longmapsto\:y = 2$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 2 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

Consider,

$\rm :\longmapsto\:y = - x + 1 \: \: when \: x < 0$

Substituting 'x = - 1' in the given equation, we get

$\rm :\longmapsto\:y = 1 + 1$

$\rm :\longmapsto\:y = 2$

Substituting 'x = - 2' in the given equation, we get

$\rm :\longmapsto\:y = 2 + 1$

$\rm :\longmapsto\:y = 3$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf 2 \\ \\ \sf - 2 & \sf 3 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.