An equation for the tangent line to the curve $y = \left( x ^ { 2 } + x - 1 \right) ^ { 3 }$ at the point where x = 1 is:
A) y = 9x
B) y = 9x - 1
C) y = 9x - 8
D) y = 2x + 1

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the curve at a specific point, we first need to find the derivative of the given function to determine the slope of the tangent line at that point. The given function is $y = (x^2 + x - 1)^3$. Using the chain rule, the derivative $y'$ is:$$y' = 3(x^2 + x - 1)^2 \cdot (2x + 1)$$Substituting $x = 1$ into the derivative to find the slope at $x = 1$:$$y'(1) = 3(1^2 + 1 - 1)^2 \cdot (2\cdot1 + 1) = 3(1)^2 \cdot 3 = 9$$So, the slope of the tangent line at $x = 1$ is 9. To find the equation of the tangent line, we use the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point on the line. First, we need to find $y_1$ by substituting $x = 1$ into the original equation:$$y_1 = (1^2 + 1 - 1)^3 = 1^3 = 1$$Thus, the point is $(1, 1)$, and the slope $m = 9$. Substituting these values into the point-slope form:$$y - 1 = 9(x - 1)$$$$y = 9x - 9 + 1$$$$y = 9x - 8$$Therefore, the correct equation of the tangent line is $y = 9x - 8$.

An Equation for the Tangent Line to the Curve y=(x2+x−1)3y = \left( x ^ { 2 } + x - 1 \right) ^ { 3 }y=(x2+x−1)3

An Equation for the Tangent Line to the Curve $y = \left( x ^ { 2 } + x - 1 \right) ^ { 3 }$