Question

The line tangent to the graph of g(x) = x^{3}-4x+1 at the point (2, 1) is given by the formula

Answer 1

well, about A and D, I just plugged the values on the slope formula of

$\bf \begin{array}{llll} g(x)=x^3-4x+1\\ L(x) = 8(x-2)+1 \end{array} \qquad \begin{cases} x_1=1.9\\ x_2=2.1 \end{cases}\implies \cfrac{f(b)-f(a)}{b-a}$

for A the values are 8.01 and 8.0, so indeed those "slopes" are close. $\textit{\huge \checkmark}$

for D the values are -2.25 and 8.0, so no dice on that one.

for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.

checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice. $\textit{\huge \checkmark}$

for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.

for E, do they intercept at (2,1)?  well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent. $\textit{\huge \checkmark}$

for F, we know the slope of the line L(x) is 8, is g'(2) = 8?  let's check

recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice. $\textit{\huge \checkmark}$