Answer
(a) $(4,0)$, $(0,\pm2)$, (b) symmetric with respect to the x-axis. (c) See graph.
Work Step by Step
(a) Given $x+y^2=4$, we can find the x-intercept(s) (let y=0) as $(4,0)$, y-intercept(s) (let x=0) as $(0,\pm2)$,
(b) To test symmetry, replace $(x,y)$ with $(x,-y)$ (x-axis symmetry), or $(-x,y)$ (y-axis symmetry), or $(-x,-y)$ (origin symmetry), and due to the $y^2$ term, we can find the equation is symmetric with respect to the x-axis.
(c) See graph.
