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 - 8
B) y = 9x
C) y = 2x + 1
D) y = 9x - 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 $3(x^2 + x - 1)^2 \cdot (2x + 1)$. Substituting $x = 1$ into the derivative gives $y'(1) = 3(1^2 + 1 - 1)^2 \cdot (2\cdot1 + 1) = 3 \cdot 1^2 \cdot 3 = 9$, which is the slope of the tangent line at $x = 1$. To find the y-intercept of the tangent line, we use the point on the curve where $x = 1$. Substituting $x = 1$ into the original equation gives $y = (1^2 + 1 - 1)^3 = 1$. The equation of the tangent line in slope-intercept form is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. We have $m = 9$ and a point $(1, 1)$. Substituting these into the equation gives $1 = 9(1) + b$, solving for $b$ gives $b = -8$. Therefore, the 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 $y = \left( x ^ { 2 } + x - 1 \right) ^ { 3 }$