Answer
$a)$ $x$-intercepts: $\pm2;$ $y$-intercept: None
$b)$ $x$-intercept: $\dfrac{1}{4};$ $y$-intercept: $1$
Work Step by Step
$a)$ $9x^{2}-4y^{2}=36$
To find the $x$-intercepts, set $y$ equal to $0$ and solve for $x$:
$9x^{2}-4y^{2}=36$
$9x^{2}-4(0)^{2}=36$
$9x^{2}=36$
$x^{2}=\dfrac{36}{9}$
$x^{2}=4$
$x=\pm\sqrt{4}$
$x=\pm2$
To find the $y$-intercept, set $x$ equal to $0$ and solve for $y$:
$9x^{2}-4y^{2}=36$
$9(0)^{2}-4y^{2}=36$
$-4y^{2}=36$
$y^{2}=\dfrac{36}{-4}$
$y^{2}=-9$
$y=\pm\sqrt{-9}$
Since solving for $y$ yields a complex solution, this equation does not have $y$-intercept.
$b)$ $y-2xy+4x=1$
To find the $x$-intercept, set $y$ equal to $0$ and solve for $x$:
$y-2xy+4x=1$
$0-2x(0)+4x=1$
$4x=1$
$x=\dfrac{1}{4}$
To find the $y$-intercept, set $x$ equal to $0$ and solve for $y$:
$y-2xy+4x=1$
$y-2(0)y+4(0)=1$
$y=1$
