Question

write an equation of the line tangent to the graph of (g(x)=x^{3}-6x) at the point where (x = 2).

Explanation:

Step1: Find the derivative of the function

The derivative of $y = g(x)=x^{3}-6x$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $g^\prime(x)=3x^{2}-6$.

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $g^\prime(x)$: $g^\prime(2)=3\times2^{2}-6=3\times4 - 6=12 - 6 = 6$. The slope of the tangent line $m = 6$.

Step3: Find the $y$ - coordinate of the point on the curve

Substitute $x = 2$ into $g(x)$: $g(2)=2^{3}-6\times2=8 - 12=-4$. The point on the curve is $(2,-4)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(2,-4)$ and $m = 6$. So $y+4 = 6(x - 2)$.

Step5: Simplify the equation

Expand the right - hand side: $y+4=6x-12$. Then $y=6x-16$.

Answer:

$y = 6x-16$