An equation for the tangent line to the curve $y = \left( 7 x ^ { 2 } + x - 1 \right) ^ { 5 }$ at the point where x = 0 is
A)y = 5x + 1
B)y = 5x - 1
C)y = 10x - 1
D)y = 10x + 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, which will give us the slope of the tangent line at any point x. The given function is $y = (7x^2 + x - 1)^5$. Using the chain rule, the derivative $y'$ is:$$y' = 5(7x^2 + x - 1)^4 \cdot (14x + 1)$$At $x = 0$, we substitute $x = 0$ into the derivative to find the slope of the tangent line:$$y'(0) = 5(7(0)^2 + 0 - 1)^4 \cdot (14(0) + 1) = 5(-1)^4 \cdot 1 = 5$$So, the slope of the tangent line at $x = 0$ is 5. 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 of tangency. Since we're looking at the point where $x = 0$, we need to find $y$ at $x = 0$ from the original equation:$$y = (7(0)^2 + 0 - 1)^5 = (-1)^5 = -1$$Thus, the point of tangency is $(0, -1)$. Using the point-slope form with $m = 5$, $x_1 = 0$, and $y_1 = -1$:$$y - (-1) = 5(x - 0)$$$$y + 1 = 5x$$$$y = 5x - 1$$Therefore, the equation of the tangent line is $y = 5x - 1$, which corresponds to choice B.

An Equation for the Tangent Line to the Curve y=(7x2+x−1)5y = \left( 7 x ^ { 2 } + x - 1 \right) ^ { 5 }y=(7x2+x−1)5

An Equation for the Tangent Line to the Curve $y = \left( 7 x ^ { 2 } + x - 1 \right) ^ { 5 }$