Question

In Exercises 71-74. Find the x- and y-intercepts of the graph of the equation.

(a) $9x^2-4y^2=36$

(b) $y-2xy+4x=1$

Short answer

(a) The x-intercepts are $-2\text{and 2}$ and y-intercept does not exist.

(b) The x-intercept is $\frac{1}{4}$ and y-intercept is $1$.

Step-by-step solution

Step 1. Apply the concept of intercepts

The x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

The y-intercepts are the points on y-axis where the graph of the equation intersects the y-axis.

Step 2. Compute the x-intercept

Consider the equation $9x^2-4y^2=36$.

To find the x-intercept, substitute $y=0$in the equation $9x^2-4y^2=36$.

$\begin{array}{c}9x^2-4{\left(0\right)}^2=36\\ 9x^2=36\\ x^2=4\\ x=-2,2\end{array}$

Therefore, x-intercepts are $-2\text{and 2}$.

Step 3. Compute the y-intercept

Consider the equation $9x^2-4y^2=36$.

To find the y-intercept, substitute $x=0$in the equation $9x^2-4y^2=36$.

$\begin{array}{c}9{\left(0\right)}^2-4y^2=36\\ -4y^2=36\\ y^2=-9\end{array}$

Therefore, no y-intercepts.

Hence, the x-intercepts are $-2\text{and 2}$ and y-intercept does not exist.

Step 1. Apply the concept of intercepts

The x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

The y-intercepts are the points on y-axis where the graph of the equation intersects the y-axis.

Step 2. Compute the x-intercept

Consider the equation $y-2xy+4x=1$.

To find the x-intercept, substitute $y=0$in the equation $y-2xy+4x=1$.

$\begin{array}{c}0-2x\left(0\right)+4x=1\\ 4x=1\\ x=\frac{1}{4}\end{array}$

Therefore, x-intercept is $\frac{1}{4}$.

Step 3. Compute the y-intercept

Consider the equation $y-2xy+4x=1$.

To find the y-intercept, substitute $x=0$in the equation $y-2xy+4x=1.$

$\begin{array}{c}y-2\left(0\right)y+4\left(0\right)=1\\ y=1\end{array}$

Therefore, y-intercept is $1$.

Hence, the x-intercept is $\frac{1}{4}$ and y-intercept is $1$.