How do you plot the following graph $ x=\sqrt{y-1} $ ?

Hint: In this question, we need to plot the graph for the given equation $ x=\sqrt{y-1} $ . For this, we will take square both sides and graph for $ y={{x}^{2}}+1 $ . We will suppose some points for x and find values of y. Using the points, we will plot the graph for $ y={{x}^{2}}+1 $ which will be a parabola.

Complete step by step answer:
Here we are given the equation of the function as $ x=\sqrt{y-1} $ .
We need to draw its graph. For this let us first simplify the equation,
Squaring both sides we get $ {{x}^{2}}={{\left( \sqrt{y-1} \right)}^{2}}\Rightarrow {{x}^{2}}=y+1 $ .
Taking 1 to the other side we get $ y={{x}^{2}}+1 $ .
We need to draw a graph for this equation.
As $ {{x}^{2}} $ is positive and it is in the quadratic form so our graph shape will be an upward parabola $ \left( \bigcup \right) $ . In general form, $ y=a{{x}^{2}}+bx+c $ the bx part of the equation shifts graph left or right but here we do not have, so our graph is central about the y-axis. Since the C part is +1 so the general parabola will shift upward.
Let us find some points to plot the graph,
Taking x = 0 we get $ y={{\left( 0 \right)}^{2}}+1\Rightarrow y=1 $ .
Taking x = 1 we get $ y={{\left( 1 \right)}^{2}}+1\Rightarrow y=2 $ .
Taking x = 2 we get $ y={{\left( 2 \right)}^{2}}+1\Rightarrow y=5 $ .
Taking x = 3 we get $ y={{\left( 3 \right)}^{2}}+1\Rightarrow y=10 $ .
Taking x = -1 we get $ y={{\left( -1 \right)}^{2}}+1\Rightarrow y=2 $ .
Taking x = -2 we get $ y={{\left( -2 \right)}^{2}}+1\Rightarrow y=5 $ .
Taking x = -3 we get $ y={{\left( -3 \right)}^{2}}+1\Rightarrow y=10 $ .
So our table of points that we will plot looks like this,




Now, let us draw the y-axis as a vertical line and x-axis as the horizontal line and mark the variables O, X, Y. Let us mark relevant points for the x and y-axis. Now let us plot all the points.
After this, let us draw a smooth curve passing through these points. This graph is our final answer.


Note:
Students should always try to find a point for the negative x-axis too because as observed earlier, this parabola is centered along the y-axis. Note that for any value of x, y cannot be negative because the equation involves $ {{x}^{2}} $ only. Make sure to make a smooth curve for more accuracy, students can plot more points.