Answer
(a) $(\pm\sqrt 5,0)$, $(0,-5)$, (b) symmetric with respect to the y-axis. (c) See graph.
Work Step by Step
(a) Given $y=x^2-5$, we can find the x-intercept(s) (let y=0) as $(\pm\sqrt 5,0)$, y-intercept(s) (let x=0) as $(0,-5)$,
(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 we can find the function is symmetric with respect to the y-axis due to the $x^2$ term.
(c) See graph.
