Answer
(a) $(\frac{3}{2}, 4)$
(b) $(\frac{3}{2}, -4)$
Work Step by Step
(a)
In an even function, for every point $(x,y)$ there also exists point $(-x,y)$
Since we know $(-\frac{3}{2}, 4)$ exists, $(\frac{3}{2}, 4)$ also has to exist.
(b)
In an odd function, for every point $(x,y)$ there also exists point $(-x,-y)$
Since we know $(-\frac{3}{2}, 4)$ exists, $(\frac{3}{2}, -4)$ also has to exist.
