Spatial Exercise IV$\mathrm{.}4\mathrm{.}1(25|89).$ This is the function pow, and pow, tangentially at $(1,1)$:$($a$)$ Why does the requirement that pow, and L intersect at $(1,1)$ imply that the difference pow, $-$ L has a zero when a is $1?($b$)$ What would the intersection look like if the degree of the intersection were $1?($c$)$ Is it possible that the intersection in this case is greater than $2?($d$)$ Write pow$($z$)$ as pow$2(1+($z $-1))$ and expand this second expression in powers that involve $($r $-1)$ to obtain a formula for pow, $($r$)$ that involves a linear term and a degree $2$ term in the argument $($a $-1).$Tags: IV$\mathrm{.}4\mathrm{.}1\mathrm{.}1,$ IV$\mathrm{.}4\mathrm{.}1\mathrm{.}2,$ IV$\mathrm{.}4\mathrm{.}1\mathrm{.}3$