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?
I agree with your dy/dx and your solution for x = 1/2 and y = 1/4
so let's try x = 1/2
1 - 2 + 12y^2 - 6y = 10
12y^2 - 6y - 11 = 0
x = (6 ± √564)/24
= (3±√141)/12
so the two points with horizontal tangents are (1/2,(3+√141)/12) and
(1/2,(3-√141)/12)
Just repeat for the other value carefully substituting and solving using the quadratic equation
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 obtained the derivative as (8x-4)/(-24y+6).
To find the points of the horizontal tangent line, you set the derivative equal to zero:
(8x-4) = 0
Solving this equation for x:
8x = 4
x = 4/8
x = 1/2
Therefore, the x-coordinate of the point of horizontal tangency is 1/2.
To find the points of the vertical tangent line, you set the derivative equal to zero:
(-24y + 6) = 0
Solving this equation for y:
-24y = -6
y = -6/-24
y = 1/4
Therefore, the y-coordinate of the point of vertical tangency is 1/4.
Now, you need to substitute these values of x and y back into the original equation to find the actual points of tangency:
For the horizontal tangent line (x = 1/2):
4(1/2)^2 - 4(1/2) + 12y^2 - 6y = 10
1 - 2 + 12y^2 - 6y = 10
12y^2 - 6y - 11 = 0
Since this quadratic equation does not factor easily, you can use the quadratic formula to find the values of y:
y = (-(-6) ± √((-6)^2 - 4(12)(-11))) / (2(12))
Simplifying this expression will give you the values of y for the horizontal tangent line. Plug these values back into the original equation to find the corresponding x-coordinates.
For the vertical tangent line (y = 1/4):
4x^2 - 4x + 12(1/4)^2 - 6(1/4) = 10
4x^2 - 4x + 3/4 - 3/2 = 10
4x^2 - 4x - 7/4 = 0
Again, since this quadratic equation does not factor easily, you can use the quadratic formula to find the values of x:
x = (-(-4) ± √((-4)^2 - 4(4)(-7/4))) / (2(4))
Simplifying this expression will give you the values of x for the vertical tangent line. Plug these values back into the original equation to find the corresponding y-coordinates.
It is important to carefully solve these equations to obtain accurate points of tangency.