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Quiz \#02 (Derivative \& Graphing) ' 1. Find the equation of the tangent line to the curve. \( y=x^{4}-3 x^{2}+2 \quad \) at the point when \( x=1 \)

Ask by Horton Pope. in Canada
Jan 23,2025

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Answer

The equation of the tangent line at \( x = 1 \) is \( y = -2x + 2 \).

Solution

To find the equation of the tangent line to the curve \( y = x^4 - 3x^2 + 2 \) at the point where \( x = 1 \), follow these steps: ### 1. **Find the Point of Tangency** First, determine the y-coordinate when \( x = 1 \): \[ y = (1)^4 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \] So, the point of tangency is **\( (1, 0) \)**. ### 2. **Find the Derivative** Next, find the derivative \( \frac{dy}{dx} \) to determine the slope of the tangent line: \[ y = x^4 - 3x^2 + 2 \] \[ \frac{dy}{dx} = 4x^3 - 6x \] ### 3. **Evaluate the Derivative at \( x = 1 \)** Substitute \( x = 1 \) into the derivative to find the slope \( m \): \[ m = 4(1)^3 - 6(1) = 4 - 6 = -2 \] ### 4. **Write the Equation of the Tangent Line** Using the point-slope form of a line: \[ y - y_1 = m(x - x_1) \] Substitute \( m = -2 \) and \( (x_1, y_1) = (1, 0) \): \[ y - 0 = -2(x - 1) \] \[ y = -2x + 2 \] ### **Final Answer** The equation of the tangent line is: \[ \boxed{y = -2x + 2} \]

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The Deep Dive

To find the equation of the tangent line at \( x = 1 \), first, we need to calculate the derivative of the function \( y = x^4 - 3x^2 + 2 \). The derivative, \( y' = 4x^3 - 6x \). Now, plug in \( x = 1 \) to find the slope: \( y'(1) = 4(1)^3 - 6(1) = 4 - 6 = -2 \). Next, calculate the y-coordinate by substituting \( x = 1 \) back into the original equation: \( y(1) = 1^4 - 3(1)^2 + 2 = 1 - 3 + 2 = 0 \). Now we have the point (1,0) and slope -2. The equation of the tangent line is \( y - 0 = -2(x - 1) \), which simplifies to \( y = -2x + 2 \). So, the equation of the tangent line is \( y = -2x + 2 \).

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