Answer
a. YES
b. If $x=0$, $f(0)=\frac{1}{2}$; The point on the graph is $(0,\frac{1}{2})$.
c. $(0, \frac{1}{2})$ and $(2, \frac{1}{2})$
d. All real numbers except $x=-4$
e. no $x$ intercept
f. $(0, \frac{1}{2})$
Work Step by Step
(a) The point $\left(1,\frac{3}{5}\right)$ is on the graph of $f$ if its coordinates satisfy the equation.
$f(x)=\dfrac{x^2 +2}{x+4}$
$\dfrac{3}{5}=\dfrac{1^2 +2}{1+4}$
$\dfrac{3}{5}\stackrel{\checkmark}{=}\dfrac{3}{5}$
Thus, the $\left(1,\frac{3}{5}\right)$ is on the graph of $f$.
(b) Substitute $0$ to $x$:
$f(x)=\dfrac{x^2 +2}{x+4}$
$f(0)=\dfrac{0^2 +2}{0+4}$
$f(0)=\dfrac{1}{2}$
Therefore, the point on the graph is $\left(0,\frac{1}{2}\right)$
(c) Substitute $\frac{1}{2}$ to $f(x)$ then solve for $x$:
$f(x)=\dfrac{x^2 +2}{x+4}$
$\dfrac{1}{2}=\dfrac{x^2 +2}{x+4}$
$x+4=2x^2 +4$
$4-4=2x^2-x$
$0=2x^2-x$
$0=x(2x-1)$
$x=0, \frac{1}{2}$
Therefore, the points on the graph are $\left(0, \frac{1}{2}\right)$ and $\left(2, \frac{1}{2}\right)$.
(d) The given function is defined for all real numbers except $x=-4$ as it makes the denominator $0$.
Thus, the domain is all real numbers except $x=-4$
(e) To find the $x$-intercept/s, set $y=0$ then solve for $x$.
$y=\dfrac{x^2 +2}{x+4}$
$0=\dfrac{x^2 +2}{x+4}$
The equation has no solution since no value of $x$ will make $\dfrac{x^2+2}{x+4}$ equal to $0$.
Therefore, there is no $x$ intercept in the given function .
(f) To find the $y$-intercept, set $x=0$ then solve for $y$.
$y=\dfrac{x^2 +2}{x+4}$
$y=\dfrac{0^2 +2}{0+4}$
$y=\dfrac{1}{2}$
Therefore, the $y$ intercept is $\left(0, \frac{1}{2}\right)$
