Question

given the function f(x)=x^3 - 9x^2 + 23x - 18, 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 -1. answer attempt 1 out of 2 two solutions and

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=x^{3}-9x^{2}+23x - 18$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=3x^{2}-18x + 23$.

Step2: Set the derivative equal to the given slope

We want the slope of the tangent line to be $-1$, so we set $f'(x)=-1$. That gives us the equation $3x^{2}-18x + 23=-1$.

Step3: Rearrange the equation to standard quadratic form

Rearrange $3x^{2}-18x + 23=-1$ to $3x^{2}-18x+24 = 0$. Divide through by 3 to simplify: $x^{2}-6x + 8=0$.

Step4: Solve the quadratic equation

Factor the quadratic equation $x^{2}-6x + 8=(x - 2)(x - 4)=0$. Using the zero - product property, we get $x=2$ or $x = 4$.

Step5: Find the corresponding y - values

When $x = 2$, $y=f(2)=2^{3}-9\times2^{2}+23\times2-18=8-36 + 46-18=0$.
When $x = 4$, $y=f(4)=4^{3}-9\times4^{2}+23\times4-18=64-144+92-18=-6$.

Answer:

$(2,0)$ and $(4,-6)$