Find equations for two lines that are both tangent to the curve y=x^3 - 3x^2 + 3x - 9 and parallel to the line 3x-y=13.

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.

Discover how to find two tangent lines to the curve y = x^3 - 3x^2 + 3x - 9 that are parallel to 3x - y = 13. This step-by-step solution covers concepts such as derivatives, parallel lines, and tangent equations. Ideal for high school students (Grades 10-12) learning about calculus and analytical geometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation