Question

The curve \(y = x^3 - 2x^2 + 3x - 4\) intersects the horizontal line \(y = -2\) at the point \(P(h,k)\). If the tangent drawn to this curve at \(P\) meets the X-axis at \((x_1, y_1)\), then \(x_1 =\)

• A. 1
• B. 2
• C. 3
• D. -3

Hint:

Check each root's feasibility and ensure arithmetic operations are precise, especially in polynomial equations.

Answer 1

The Correct Option is B

To solve the problem, we need to find the point \( P(h, k) \) where the curve \( y = x^3 - 2x^2 + 3x - 4 \) intersects the horizontal line \( y = -2 \). Then, we will determine the equation of the tangent to the curve at point \( P \) and find where this tangent intersects the X-axis. Step 1: Find the point \( P(h, k) \) Set \( y = -2 \) equal to the equation of the curve: \[ -2 = x^3 - 2x^2 + 3x - 4 \] Rearrange the equation: \[ x^3 - 2x^2 + 3x - 2 = 0 \] We need to find the real root of this cubic equation. Let's test \( x = 1 \): \[ 1 - 2 + 3 - 2 = 0 \] So, \( x = 1 \) is a root. Therefore, \( h = 1 \) and \( k = -2 \). Thus, \( P(1, -2) \). Step 2: Find the slope of the tangent at \( P \) First, compute the derivative of the curve: \[ \frac{dy}{dx} = 3x^2 - 4x + 3 \] Evaluate the derivative at \( x = 1 \): \[ \frac{dy}{dx}\bigg|_{x=1} = 3(1)^2 - 4(1) + 3 = 3 - 4 + 3 = 2 \] So, the slope \( m \) of the tangent at \( P \) is 2. Step 3: Find the equation of the tangent line Using the point-slope form: \[ y - k = m(x - h) \] Substitute \( h = 1 \), \( k = -2 \), and \( m = 2 \):
\includegraphics[width=0.5\linewidth]{68I1.png}
Step 4: Find the X-intercept of the tangent line Set \( y = 0 \) to find the X-intercept:
\includegraphics[width=0.5\linewidth]{68I2.png}
Therefore, the tangent line intersects the X-axis at \( (2, 0) \), so \( x_1 = 2 \). Final Answer: \[ \boxed{2} \]