Choose a correct statement about the slope and equation of the tangent line to the curve

f(x)=x3−3x2−x+3  at the point  (x=2,y=−3).

 A.

Slope is −1 and equation is x−y+1=0

 B.

Slope is undefined and equation is x=2.

 C.

Slope is −3 and equation is y=3−3x

 D.

Slope is 0 and equation is y=−3

 E. No correct answer.

To find the correct statement about the slope and the equation of the tangent line to the curve \( f(x) = x^3 - 3x^2 - x + 3 \) at the point \((x = 2, y = -3)\), let's follow these steps:

### Step 1: Differentiate the function to find the slope of the tangent line

The slope of the tangent line at a point on the curve is given by the derivative of the function at that point. Let's differentiate \( f(x) \):

\[
f(x) = x^3 - 3x^2 - x + 3
\]

\[
f'(x) = \frac{d}{dx}(x^3 - 3x^2 - x + 3) = 3x^2 - 6x - 1
\]

Now, substitute \( x = 2 \) into \( f'(x) \) to find the slope at the point \((2, -3)\):

\[
f'(2) = 3(2)^2 - 6(2) - 1 = 12 - 12 - 1 = -1
\]

So, the slope of the tangent line at \( x = 2 \) is \(-1\).

### Step 2: Find the equation of the tangent line

The equation of the tangent line can be written in point-slope form:

\[
y - y_1 = m(x - x_1)
\]

where \((x_1, y_1) = (2, -3)\) and the slope \( m = -1 \).

Substituting these values:

\[
y - (-3) = -1(x - 2)
\]

\[
y + 3 = -x + 2
\]

\[
x + y + 1 = 0
\]

or, equivalently:

\[
x - y + 1 = 0
\]

### Step 3: Evaluate the given options

- **Option A: Slope is \(-1\) and equation is \(x - y + 1 = 0\)**  
  This is correct based on our calculations.

- **Option B: Slope is undefined and equation is \(x = 2\)**  
  This is incorrect because the slope is defined and is \(-1\).

- **Option C: Slope is \(-3\) and equation is \(y = 3 - 3x\)**  
  This is incorrect as we found the slope to be \(-1\).

- **Option D: Slope is \(0\) and equation is \(y = -3\)**  
  This is incorrect since the slope is not \(0\).

- **Option E: No correct answer.**  
  This is incorrect because Option A is correct.

### Conclusion

The correct answer is:

**A. Slope is \(-1\) and equation is \(x - y + 1 = 0\).**

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Would you like more details or have any questions?

### Related Questions:

1. How do you find the tangent line to a curve at a specific point?
2. What are the critical points of the function \(f(x) = x^3 - 3x^2 - x + 3\)?
3. How do you find where the function \(f(x) = x^3 - 3x^2 - x + 3\) is increasing or decreasing?
4. What is the concavity of the function \(f(x) = x^3 - 3x^2 - x + 3\) at the point \(x = 2\)?
5. How do you find the second derivative of the function \(f(x) = x^3 - 3x^2 - x + 3\) and what does it represent?

**Tip:** Always check your calculations carefully when determining the slope and equation of tangent lines; small errors can lead to incorrect conclusions!

Learn how to find the slope and equation of the tangent line to the curve f(x)=x^3-3x^2-x+3 at the point (x=2, y=-3). Detailed step-by-step solution includes differentiation, point-slope form, and evaluation of options. Suitable for advanced high school students studying calculus.

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