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 \( x=1 \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
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 \( y = -2x + 2 \).