Question

consider the equation and the given point.
$f(x)=(x^{3}+9x - 1)(x - 2),\\ (1,-9)$
(a) find an equation of the tangent line to the graph of $f$ at the given point.
$y = square$

Explanation:

Step1: Use the product - rule to find the derivative

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=x^{3}+9x - 1$ and $v=x - 2$. Then $u^\prime = 3x^{2}+9$ and $v^\prime=1$. So $f^\prime(x)=(3x^{2}+9)(x - 2)+(x^{3}+9x - 1)\times1$.

Step2: Expand the derivative

Expand $(3x^{2}+9)(x - 2)+(x^{3}+9x - 1)$:
\[

$$\begin{align*} &(3x^{2}+9)(x - 2)+(x^{3}+9x - 1)\\ =&3x^{3}-6x^{2}+9x - 18+x^{3}+9x - 1\\ =&4x^{3}-6x^{2}+18x - 19 \end{align*}$$

\]

Step3: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$: $f^\prime(1)=4\times1^{3}-6\times1^{2}+18\times1 - 19=4 - 6+18 - 19=-3$. The slope of the tangent line $m=-3$.

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,-9)$ and $m=-3$.
\[

$$\begin{align*} y-(-9)&=-3(x - 1)\\ y + 9&=-3x+3\\ y&=-3x - 6 \end{align*}$$

\]

Answer:

$y=-3x - 6$