# Find the slope m of the tangent to the curve y = 3 + 4x^2 - 2x^3 at the point where x = a.

## Question:

Find the slope {eq}m {/eq} of the tangent to the curve {eq}y = 3 + 4x^2 - 2x^3 {/eq} at the point where {eq}x = a {/eq}.

## Tangent Line of a Function

Tangent line of a function is found using the equation {eq}y - y_o = y'(x_o)(x - x_o) {/eq}, where:

- {eq}x_o {/eq} is the x-coordinate of the point through which the tangent line passes,

- {eq}y_o {/eq} is the x-coordinate of the point through which the tangent line passes, also found by {eq}y(x_o) {/eq},

- {eq}y'(x_o) {/eq} is the value of the derivative at point {eq}x_o {/eq}.

## Answer and Explanation:

Our problem requires us to find the tangent line at point {eq}x_o = a. {/eq} We can readily identify the y-coordinate at this point by direct substitution of **a** into the function:

{eq}y_o = y(x_o) = 3 + 4a^2 - 2a^3. {/eq}

Now, let's find the derivative of the given function:

{eq}y'(x) = \frac{dy}{dx} = \frac{d}{dx} (3 + 4x^2 - 2x^3) = 8x - 6x^2 = 2x(4 - 3x). {/eq}

Substituting the value **a** into the expression of **y'(x)**, we'll obtain the slope:

{eq}y'(x_o) = 2a(4 - 3a). {/eq}

Currently, we have all the terms required to write the tangent line equation:

{eq}y - y_o = y'(x_o)(x - x_o),\\ y - (3 + 4a^2 - 2a^3) = 2a(4 - 3a)(x - a). {/eq}

Simplify this equation:

{eq}y - (3 + 4a^2 - 2a^3) = (8a - 6a^2)(x - a),\\ y - 3 - 4a^2 + 2a^3 = 8ax - 8a^2 - 6a^2x + 6a^3,\\ y = 8ax - 8a^2 - 6a^2x + 6a^3 - 2a^3 + 4a^2 + 3,\\ y = (8a - 6a^2)x - 4a^2 + 4a^3 + 3,\\ y = 2a(4 - 3a)x + 4a^2(a - 1) + 3. {/eq}

As a result, the slope of the tangent to the given curve at point **x = a** is the coefficient adjacent to **x**:

{eq}\boxed{m = 2a(4 - 3a)} {/eq}.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11