Questions: For (y=2 x^3+6 x^2+5 x+5) [ fracd yd x= ] What is the slope of the curve at (x=1) ? The equation of the tangent line at (x=1) is:

$For (y=2 x^3+6 x^2+5 x+5) [ fracd yd x= ] What is the slope of the curve at (x=1) ? The equation of the tangent line at (x=1) is:$

Transcript text: For $y=2 x^{3}+6 x^{2}+5 x+5$ \[ \frac{d y}{d x}= \] $\square$ What is the slope of the curve at $x=1$ ? $\square$ The equation of the tangent line at $x=1$ is: $\square$

Solution

Solution Steps

To solve the given problem, we need to follow these steps:

Differentiate the function $ y = 2x^3 + 6x^2 + 5x + 5 $ to find $\frac{dy}{dx}$.
Evaluate the derivative at $ x = 1 $ to find the slope of the curve at that point.
Find the equation of the tangent line at $ x = 1 $ using the point-slope form of the line equation.

Step 1: Differentiate the Function

Given the function $ y = 2x^3 + 6x^2 + 5x + 5 $, we need to find its derivative with respect to $ x $.

\[ \frac{dy}{dx} = 6x^2 + 12x + 5 \]

Step 2: Evaluate the Derivative at $ x = 1 $

To find the slope of the curve at $ x = 1 $, we substitute $ x = 1 $ into the derivative.

\[ \left. \frac{dy}{dx} \right|_{x=1} = 6(1)^2 + 12(1) + 5 = 23 \]

Step 3: Find the $ y $-coordinate at $ x = 1 $

We substitute $ x = 1 $ into the original function to find the corresponding $ y $-coordinate.

\[ y(1) = 2(1)^3 + 6(1)^2 + 5(1) + 5 = 18 \]

Step 4: Equation of the Tangent Line at $ x = 1 $

Using the point-slope form of the line equation $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $(x_1, y_1)$ is the point of tangency:

\[ y - 18 = 23(x - 1) \]