# Find an equation of the tangent line to the given curve at the given point. y = 3x^4 - x^3 - x^2...

## Question:

Find an equation of the tangent line to the given curve at the given point.

{eq}y = 3x^4 - x^3 - x^2 + 1, \ \ \ (1, 2) {/eq}

## Tangent line:

The line that touches a curve at only one point is known as a tangent line to that curve, the general solution of that tangent line at a point {eq}(x_0, y_0) {/eq} is given by,

$$y - y_0 = y'_{(x_0, y_0)} (x - x_0) $$

{eq}y' {/eq} is the derivative of the given curve.

## Answer and Explanation:

We have,

$$y = 3x^4 - x^3 - x^2 + 1 $$

Differentiate the given function with respect to {eq}x {/eq},

$$\displaystyle \begin{align*} y' &= \frac {d}{dx} \left [ 3x^4 - x^3 - x^2 + 1 \right ] \\[0.3 cm] &= \frac {d}{dx} \left [ 3x^4 \right ] - \frac {d}{dx} \left [ x^3 \right ] - \frac {d}{dx} \left [ x^2 \right ] + \frac {d}{dx} \left [ 1 \right ] \\[0.3 cm] &= 3\left ( 4x^3 \right ) - 3x^2 - 2x + 0 \\[0.3 cm] &= 12x^3 - 3x^2 - 2x \\[0.3 cm] \text {At} \ x = 1 \ &. \\[0.3 cm] y' &= 12(1)^3 - 3(1)^2 - 2(1) \\[0.3 cm] &= 12 - 3 - 2 \\[0.3 cm] &= 12 - 5 \\[0.3 cm] y' &= 7 \end{align*} $$

The equation of the tangent line to the curve {eq}y = 3x^4 - x^3 - x^2 + 1 {/eq} at the point {eq}(1, 2) {/eq} is given by,

$$\displaystyle \begin{align*} y - 2 &= (7)(x - 1) \\[0.3 cm] y - 2 &= 7x - 7 \\[0.3 cm] y &= 7x - 7 + 2 \\[0.3 cm] y &= 7x - 5 \\[0.3 cm] \end{align*} $$

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from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11