Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,356 questions

17,804 answers

1,438 comments

35,964 users

Find equation for line tangent - Basic Calculus

0 votes

The equation for the line tangent y=4-8x^2 at (5,-196)

asked Dec 25, 2012 in CALCULUS by linda Scholar
reshown Dec 25, 2012 by moderator

2 Answers

+1 vote
Given equation is y=4-8x^2

The tangent of the given equation is derivative of the given equation

 dy/dx = d(4-8x^2)/dx

          = d(4)/dx-d(8x^2)/dx

          = 0-16x

dy/dx  = -16x

Substitute (5,-196) in tangent line

= -16(5)

= -80.

This is the equation for the tangent line.
answered Dec 25, 2012 by friend Mentor

The tangent line equation  is y = - 80x + 204.

0 votes

The curve is y =  4 - 8x2 and the point is (5, -196).

Differentiate the curve with respect to ' x '.

y ' = - 16x.

When, x = 5, y ' = - 16 * 5 = -80.

y ' = - 80.

This is the slope (m ) of the tangent line to the implicit curve at (5, - 196).

 

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Slope (m) = - 80.

Now, the tangent line equation is y = - 80x + b.

 

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (5, - 196).

- 196 = (- 80)(5) + b

b = - 196 + 400

b = 204.

The tangent line equation  is y = - 80x + 204.

answered Jun 21, 2014 by lilly Expert

Related questions

...