# The equation of the tangent and parallel to the line

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.1, Problem 53E
To determine

## To find:The equation of the tangent and parallel to the line

Expert Solution

The equation of the both line is y9=12(x2),y+7=12(x+2)

### Explanation of Solution

Given:

The tangent to the curve

y=1+x3

Parallel to the line

12xy=1y=12x1

Concept used:

The equation is in slope −intercept form, y=mx+c

An equation for the line through the point (x1,y1) with slope m is

yy1=m(xx1)

Calculation:

The function

y=1+x3.......................(1)

The derivative of a function

y=f(x)y=f(x)=dydx=m

Differentiating the equation (1) with respect to x

y=1+x3y=3x2........................(2)

The derivative is slope of the tangent line so in order to the slope of the tangent line

The derivative of constant is zero

y=12x1y=12.............................(3)

The equation is in slope −intercept form

The slope of the line is =12

From equation (2) and equation (3)

3x2=12x2=4x=±2

The coordinates of points on each line are (2,9)and(2,7)

An equation for the line through the point (x1,y1) with slope m is

yy1=m(xx1)y9=12(x2)y+7=12(x+2)

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