How do you graph y=4/(x-3)+2 using asymptotes, intercepts, end behavior?

Jul 13, 2018

Vertical asymptote at $x = 3$, horizontal asymptote at
$y = 2$, x intercept is at $\left(1 , 0\right)$, y intercept is at $\left(0 , \frac{2}{3}\right)$.
End behavior : $x \to - \infty , y \to 2 \mathmr{and} x \to \infty , y \to 2$

Explanation:

$y = \frac{4}{x - 3} + 2$

Vertical asymptote at $x - 3 = 0 \mathmr{and} x = 3$

When $x \to {3}^{-} , y \to - \infty \mathmr{and} x \to {3}^{+} , y \to \infty$

Since denominator's degree is higher than that of numerator,

horizontal asymptote is at $y = 0 + 2 \mathmr{and} y = 2$

y intercept is at $x = 0 \therefore y = \frac{4}{0 - 3} + 2 = - \frac{4}{3} + 2 \mathmr{and} y = \frac{2}{3}$

x intercept is at $y = 0 \therefore 0 = \frac{4}{x - 3} + 2 \mathmr{and} - 2 = \frac{4}{x - 3}$ or

$- 2 \left(x - 3\right) = 4 \mathmr{and} 2 x = 6 - 4 \mathmr{and} 2 x = 2 \mathmr{and} x = 1$

End behavior : When $x \to - \infty , y \to 2 \mathmr{and} x \to \infty , y \to 2$

graph{(4/(x-3))+2 [-40, 40, -20, 20]} [Ans]