Answer
(a) The slope of the tangent line is $-8$.
(b) The equation of that tangent line is $$y=-8x+17$$
Work Step by Step
$$y=f(x)=9-2x^2$$
1) First we must find the slope of the tangent line to $f(x)$ at a number $a$ $(m_a)$: $$m_a=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$
$$m_a=\lim\limits_{x\to a}\frac{(9-2x^2)-(9-2a^2)}{x-a}$$
$$m_a=\lim\limits_{x\to a}\frac{-2x^2+2a^2}{x-a}$$
$$m_a=\lim\limits_{x\to a}\frac{-2(x^2-a^2)}{x-a}$$
$$m_a=\lim\limits_{x\to a}\frac{-2(x-a)(x+a)}{x-a}$$
$$m_a=\lim\limits_{x\to a}[-2(x+a)]$$
$$m_a=-2(a+a)$$
$$m_a=-4a$$
(a) At point $(2, 1)$, we have $a=2$. So, the slope of the tangent line at $(2,1)$ is $$m(2)=-4\times2=-8$$
(b) The equation of that tangent line $l$ would have the following form: $$(l): y= m(2)x+b$$$$(l): y=-8x+b$$
Now we know that $l$ is the tangent line to $f(x)$ at $(2,1)$. Therefore, the point $(2,1)$ must lie in $l$.
So, we can apply that point to find $b$: $$-8\times2+b=1$$$$b=17$$
Therefore, the equation of the tangent line at $(2,1)$ is $$(l):y=-8x+17$$
