Question

given the function $f(x)=x^{3}+6x^{2}+6x + 10$, determine all coordinate points $(x,y)$ on the graph of $f$ such that the line tangent to $f$ at $(x,y)$ has a slope of $- 3$. answer attempt 1 out of 2 two solutions and

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{3}+6x^{2}+6x + 10$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=3x^{2}+12x + 6$.

Step2: Set the derivative equal to the given slope

We want to find $x$ such that $f'(x)=-3$. So we set up the equation $3x^{2}+12x + 6=-3$.
Rearrange it to the standard quadratic form $ax^{2}+bx + c = 0$:
$3x^{2}+12x+9 = 0$. Divide through by 3 to get $x^{2}+4x + 3=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+4x + 3=(x + 1)(x + 3)=0$.
Using the zero - product property, if $(x + 1)(x + 3)=0$, then $x+1 = 0$ or $x + 3=0$.
So $x=-1$ or $x=-3$.

Step4: Find the corresponding $y$ values

When $x=-1$, $y=f(-1)=(-1)^{3}+6(-1)^{2}+6(-1)+10=-1 + 6-6 + 10=9$.
When $x=-3$, $y=f(-3)=(-3)^{3}+6(-3)^{2}+6(-3)+10=-27+54-18 + 10=19$.

Answer:

$(-1,9)$ and $(-3,19)$