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

asked Dec 25, 2012 in CALCULUS
reshown Dec 25, 2012

+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.

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

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.