Question

given the function $f(x)=x^{3}-3x^{2}-12x + 9$, 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 $f(x)=x^{3}-3x^{2}-12x + 9$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=3x^{2}-6x-12$.

Step2: Set the derivative equal to the given slope

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

Step3: Solve the quadratic equation

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

Step4: Find the corresponding $y$ - values

When $x = 3$, $y=f(3)=3^{3}-3\times3^{2}-12\times3 + 9=27-27-36 + 9=-27$.
When $x=-1$, $y=f(-1)=(-1)^{3}-3\times(-1)^{2}-12\times(-1)+9=-1-3 + 12+9=17$.

Answer:

$(-1,17)$ and $(3,-27)$