How do you find the slope of a tangent line to the graph of the function #y^2-2x-4y-1=0#, at (-2,1)?

Question

Ezekiel Alexander · Accepted Answer

Slope of the tangent is #m=-1#

The first derivative of the function gives the slope of the line at any given point.

Given -

#y^2-2x-4y-1=0#
Differentiate implicitly

#2y dy/dx-2-4dy/dx=0#
#dy/dx(2y-4)=2#
#dy/dx=2/(2y-4)#

Slope at #(-2,1)#

#m=2/(2(1)-4)=2/(-2)=-1#

Slope of the curve at point #(-2,1)# is the slope of the tangent.

Slope of the tangent is #m=-1#

Isabella Ackerman · Answer

undefined To find the slope of a tangent line to the graph of a function, we need to find the derivative of the function and evaluate it at the given point.

First, let's rearrange the given equation to express y in terms of x: 
y^2 - 4y = 2x + 1 
Completing the square, we get: 
(y - 2)^2 - 4 = 2x + 1 
(y - 2)^2 = 2x + 5

Now, we can differentiate both sides of the equation with respect to x: 
2(y - 2) * (dy/dx) = 2 
Simplifying, we have: 
(y - 2) * (dy/dx) = 1 
dy/dx = 1 / (y - 2)

To find the slope at the point (-2, 1), substitute the values into the derivative: 
dy/dx = 1 / (1 - 2) 
dy/dx = 1 / (-1) 
dy/dx = -1

Therefore, the slope of the tangent line to the graph of the function at (-2, 1) is -1.