find the points at which the graph of 4x^2 - 4x + 12y^2 -6y = 10 has a vertical and horizontal tangent line.
For the vertical only solve for y and for the horizontal only solve for x.
I got the deriviative to be:
(8x-4)/(-24y + 6).
For the horizontal, x = 1/2
For the vertical y = 1/4
I have to plug these into the original equation to get the points and solve for the rest of the equation but each time i do it, it doesn't turn out right.
Can someone check over my work and show me how to finish the problem?
Thanks
To find the points at which the graph of the equation 4x^2 - 4x + 12y^2 - 6y = 10 has a vertical and horizontal tangent line, you correctly took the derivative. The derivative you found is
(8x - 4) / (-24y + 6).
To find the points where the graph has a horizontal tangent line, you set the derivative equal to zero since the slope of a horizontal line is zero. Therefore, you have
(8x - 4) / (-24y + 6) = 0.
Simplifying this equation, you get
8x - 4 = 0.
Solving for x, you find that x = 1/2. So, the point where the graph of the given equation has a horizontal tangent line is (1/2, y).
Now, let's find the points where the graph has a vertical tangent line. To do this, you set the derivative denominator equal to zero since the slope of a vertical line is undefined. Therefore, you have
-24y + 6 = 0.
Solving for y, you get y = 1/4. So, the point where the graph has a vertical tangent line is (x, 1/4).
To find the actual points, you need to substitute these values of x and y back into the original equation. Plugging x = 1/2 into the equation 4x^2 - 4x + 12y^2 - 6y = 10, you will get an equation only in terms of y. Solve this equation for y and you will find the y-coordinate of the point. Similarly, plugging y = 1/4 into the equation will give you an equation only in terms of x. Solve this equation for x to find the x-coordinate of the point.