We are tasked with finding the equations of two lines that are tangent to the curve $y = x^3 - 3x^2 + 3x - 9$ and parallel to the line $3x - y = 13$.

Step 1: Determine the slope of the line

The equation of the given line $3x - y = 13$ can be rewritten in slope-intercept form as: $y = 3x - 13$ From this, we see that the slope $m$ of the line is 3.

Step 2: Find the slope of the tangent line to the curve

The slope of the tangent line to the curve $y = x^3 - 3x^2 + 3x - 9$ at any point $x$ is given by the derivative of the curve: $\frac{dy}{dx} = 3x^2 - 6x + 3$ We are looking for points on the curve where the slope of the tangent line is equal to 3, since the tangent lines must be parallel to the line $3x - y = 13$.

So, we set the derivative equal to 3: $3x^2 - 6x + 3 = 3$ Simplifying this: $3x^2 - 6x = 0$ $3x(x - 2) = 0$ Thus, the solutions are: $x = 0 \quad \text{or} \quad x = 2$

Step 3: Find the points of tangency

We now substitute these $x$-values into the original curve equation $y = x^3 - 3x^2 + 3x - 9$ to find the corresponding $y$-values.

• For $x = 0$: $y = 0^3 - 3(0)^2 + 3(0) - 9 = -9$ So, the point of tangency is $(0, -9)$.

• For $x = 2$: $y = 2^3 - 3(2)^2 + 3(2) - 9 = 8 - 12 + 6 - 9 = -7$ So, the point of tangency is $(2, -7)$.

Step 4: Write the equations of the tangent lines

We now write the equations of the tangent lines passing through the points $(0, -9)$ and $(2, -7)$, both with slope 3.

• For the line through $(0, -9)$: Using the point-slope form of the line equation $y - y_1 = m(x - x_1)$: $y - (-9) = 3(x - 0)$ $y = 3x - 9$

• For the line through $(2, -7)$: $y - (-7) = 3(x - 2)$ $y = 3x - 6 - 7 = 3x - 13$

Final Answer:

The equations of the two lines that are tangent to the curve and parallel to the given line are:

1. $y = 3x - 9$
2. $y = 3x - 13$

Do you want more details or have any questions?

Here are 5 related questions:

1. How do you calculate the slope of a tangent line to a curve?
2. Can a curve have more than two tangent lines parallel to a given line?
3. How does finding the derivative relate to the slope of a tangent?
4. What is the significance of using point-slope form for writing line equations?
5. How do we determine points of tangency for a curve?

Tip: The point-slope form of a line is very useful when you know the slope and a specific point on the line.