Question

The curve \( y = x^4 - 4x^3 + 4x^2 - 4 \) has tangents parallel to the x-axis at the following points \( (x, y) \)

• A. (1, 4), (-2, 2) and (0, -1)
• B. (0, -4), (2, -4) and (1, -3)
• C. (-1, 2), (2, 1) and (1, -2)
• D. (1, -4), (1, -3) and (2, -4)

Hint:

To find points where the tangent is parallel to the x-axis, set the derivative equal to zero and solve for the values of \( x \).

Answer 1

The Correct Option is B

To find the points where the tangents to the curve \( y = x^4 - 4x^3 + 4x^2 - 4 \) are parallel to the x-axis, we need to determine where the derivative of the function is zero.

First, let's find the derivative of the given function:

\(y = x^4 - 4x^3 + 4x^2 - 4\)

The derivative is:

\(\frac{dy}{dx} = \frac{d}{dx}(x^4 - 4x^3 + 4x^2 - 4) = 4x^3 - 12x^2 + 8x\)

We set the derivative equal to zero to find the critical points:

\(4x^3 - 12x^2 + 8x = 0\)

Factor out the common term:

\(4x(x^2 - 3x + 2) = 0\)

This gives us:

\(x = 0\) or \(x^2 - 3x + 2 = 0\)

Solving the quadratic equation \(x^2 - 3x + 2 = 0\):

By factoring, we get:

\((x - 1)(x - 2) = 0\)

Thus, \(x = 1\) or \(x = 2\).

So the critical points are \(x = 0, 1,\) and \(2\).

Now, find the corresponding \(y\) values for each \(x\):

• For \(x = 0\):
  \(y = 0^4 - 4 \cdot 0^3 + 4 \cdot 0^2 - 4 = -4\)
  Point: (0, -4)
• For \(x = 1\):
  \(y = 1^4 - 4 \cdot 1^3 + 4 \cdot 1^2 - 4 = 1 - 4 + 4 - 4 = -3\)
  Point: (1, -3)
• For \(x = 2\):
  \(y = 2^4 - 4 \cdot 2^3 + 4 \cdot 2^2 - 4 = 16 - 32 + 16 - 4 = -4\)
  Point: (2, -4)

Therefore, the points where the tangents to the curve are parallel to the x-axis are (0, -4), (1, -3), and (2, -4).

The correct answer is: (0, -4), (2, -4) and (1, -3).