# 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 55E
To determine

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

Expert Solution

The equation of the both line is y=13(x73)199

### Explanation of Solution

Given:

The tangent to the curve

y=x25x+4

Parallel to the line

x3y=5y=5x3

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=x25x+4.......................(1)

The derivative of a function

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

Differentiating the equation (1) with respect to x

y=x25x+4y=2x5........................(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

x3y=5y=5x3y=13.............................(3)

The equation is in slope −intercept form

The slope of the line is =13

From equation (2) and equation (3)

13=2x56x15=1x=146x=73

The coordinates of points

y=(73)25(73)+4y=499353+4y=559+4y=199

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

yy1=m(xx1)y+199=13(x73)y=13(x73)199

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!