question 8 of 8 step 1 of 1 find the equation of the tangent line to the graph of f(x) at the (x…

Explanation:

Step1: Use product - rule to find derivative

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=-3x^{2}+4x + 3$ and $v=-x^{2}+3$. Then $u^\prime=-6x + 4$ and $v^\prime=-2x$. So $f^\prime(x)=(-6x + 4)(-x^{2}+3)+(-3x^{2}+4x + 3)(-2x)$.

Step2: Expand the derivative

Expand $(-6x + 4)(-x^{2}+3)=6x^{3}-18x-4x^{2}+12$ and $(-3x^{2}+4x + 3)(-2x)=6x^{3}-8x^{2}-6x$. Then $f^\prime(x)=(6x^{3}-4x^{2}-18x + 12)+(6x^{3}-8x^{2}-6x)=12x^{3}-12x^{2}-24x + 12$.

Step3: Evaluate the derivative at $x=-1$

Substitute $x = - 1$ into $f^\prime(x)$: $f^\prime(-1)=12(-1)^{3}-12(-1)^{2}-24(-1)+12$. $=-12-12 + 24+12=12$.

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

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-1,-8)$ and $m = f^\prime(-1)=12$. $y+8=12(x + 1)$. $y+8=12x+12$. $y=12x + 4$.

Answer:

$y = 12x+4$