How do you find the points where the graph of the function #x^2+7y^2-4x-2=0# has horizontal tangents?

Question

Ezra Adams · Accepted Answer

At #(2,0.926)# and #(2,-0.926)#, graph of the function
#x^2+7y^2-4x-2=0# will have horizontal tangents.

The graph of the function #x^2+7y^2-4x-2=0# will have horizontal tangents, where its slope is zero i.e. first derivative is equal to zero. For this, factorizing #x^2+7y^2-4x-2=0# gives us #2x+14y(dy)/(dx)-4=0# or #(dy)/(dx)=(4-2x)/(14y)#, which will be zero for #(4-2x)/(14y)=0# or #4-2x=0# if #y!=0# i.e. #x=2#. Now putting #x=2# in #x^2+7y^2-4x-2=0# we get #2^2+7y^2-4xx2-2=0# or #4+7y^2-8-2=0# or #7y^2=10-4=6# or #y=sqrt(6/7)=+-0.926# Hence points are #(2,0.926)# and #(2,-0.926)# graph{x^2+7y^2-4x-2=0 [-2, 5, -2, 2]}

William Abdalla · Answer

undefined To find the points where the graph of the function x^2+7y^2-4x-2=0 has horizontal tangents, we need to find the values of x and y that satisfy the condition for horizontal tangents.

First, we differentiate the given function with respect to x to find its derivative.

d/dx (x^2+7y^2-4x-2) = 2x - 4

Next, we set the derivative equal to zero to find the critical points where the tangents are horizontal.

2x - 4 = 0

Solving this equation, we find x = 2.

Substituting this value of x back into the original equation, we can solve for y.

(2)^2 + 7y^2 - 4(2) - 2 = 0

4 + 7y^2 - 8 - 2 = 0

7y^2 - 6 = 0

Simplifying further, we have 7y^2 = 6.

Dividing both sides by 7, we find y^2 = 6/7.

Taking the square root of both sides, we get y = ±√(6/7).

Therefore, the points where the graph of the function has horizontal tangents are (2, √(6/7)) and (2, -√(6/7)).