question: find the points at which the graph of x^2 -4x + y62 -6y =12 has a vertical or horizontal tangent line.
I found the deriviative to be (x-2)/(3-y)
Then, the x value for a horizontal would be x=2 and the y value for a vertical would be y=3.
But how do i find the exact points. My teacher gave me these answers
Horizontal tangent at (2,8) and (2,-2)
Vertical tangent at (7,3) and (-3,3)
But i don't understand how she got those points. can someone show me?
Thanks
You've done the hard part, so you're probably just over-thinking it! :-)
When x=2, plug 2 into that equation of the circle, and what do you get?
x^2 -4x + y^2 -6y =12
4 - 8 + y^2 -6y = 12
y^2 -6y -16 = 0
and solve for y.
Same method works for getting the points where y=3.
To find the exact points where a graph has a horizontal or vertical tangent line, we need to equate the derivative to zero and solve for the corresponding variables.
1. Horizontal Tangent Lines:
For a horizontal tangent line, the derivative of the function with respect to y should be zero because the tangent line will have a slope of zero.
Derivative with respect to y (dy/dx):
(x-2)/(3-y) = 0
Since the denominator can't be zero, we set the numerator to zero:
x - 2 = 0
x = 2
Now we substitute the value of x back into the original equation to find the corresponding y-values.
x^2 - 4x + y^2 - 6y = 12
(2)^2 - 4(2) + y^2 - 6y = 12
4 - 8 + y^2 - 6y = 12
y^2 - 6y - 16 = 0
Factor the quadratic equation:
(y - 8)(y + 2) = 0
From this, we get two possible y-values: y = 8 or y = -2.
Therefore, the points where the graph has a horizontal tangent line are (2, 8) and (2, -2).
2. Vertical Tangent Lines:
For a vertical tangent line, the derivative of the function with respect to x should be undefined because the tangent line will have an infinite slope.
Derivative with respect to x (dy/dx):
(x-2)/(3-y) = undefined (denominator is 0)
For the denominator to be zero, 3 - y = 0.
y = 3
Now we substitute the value of y back into the original equation to find the corresponding x-values.
x^2 - 4x + y^2 - 6y = 12
x^2 - 4x + (3)^2 - 6(3) = 12
x^2 - 4x + 9 - 18 = 12
x^2 - 4x - 6 = 0
Using the quadratic formula, we find the solutions for x:
x = (4 ± √(16 + 24)) / 2
x = (4 ± √(40)) / 2
x = (4 ± 2√10) / 2
x = 2 ± √10
Therefore, the points where the graph has a vertical tangent line are (2 + √10, 3) and (2 - √10, 3).
In summary, the exact points where the graph of x^2 - 4x + y^2 - 6y = 12 has a horizontal tangent line are (2, 8) and (2, -2), while the points where it has a vertical tangent line are (2 + √10, 3) and (2 - √10, 3).