Chapter 1.2, Problem 26E

To determine

The $x-\text{intercept}$ and $y-\text{intercept}$ of the graph of the equation $y=\sqrt{2x-1}$ .

Answer to Problem 26E

The $x-\text{intercept}$ of the graph is the point $\left(x,y\right)=\left(\frac{1}{2},0\right)$

There is no $y-\text{intercept}$ of the graph

Explanation of Solution

Given:

The equation $y=\sqrt{2x-1}$

Concept Used:

The $x-\text{intercept}$ is a point on the graph where the graph intersects $x-\text{axis}$ i.e., where the $y$ coordinate is zero.

The $y-\text{intercept}$ is a point on the graph where the graph intersects $y-\text{axis}$ i.e., where the $x$ coordinate is zero.

Calculation:

Given the equation $y=\sqrt{2x-1}$

For $x-\text{intercept}$ , substituting $y=0$

i.e., $y=\sqrt{2x-1}$

i.e., $\begin{array}{c}0=\sqrt{2x-1}\\ 0^2={\left(\sqrt{2x-1}\right)}^2\text{ Squaring both sides}\\ 0=2x-1\\ x=\frac{1}{2}\end{array}$

Therefore, $x-\text{intercept}$ of the graph is the point $\left(x,y\right)=\left(\frac{1}{2},0\right)$

For $y-\text{intercept}$ , substituting $x=0$

i.e., $y=\sqrt{2x-1}$

i.e., $\begin{array}{l}y=\sqrt{2\left(0\right)-1}\\ y=\sqrt{0-1}\\ y=\sqrt{-1}\end{array}$

which is not defined for real number.

Therefore, there is no $y-\text{intercept}$ of the graph