Find all points (if any)of horizontal and vertical tangency to the curve , .

Question

Find all points (if any)of horizontal and vertical tangency to the curve  ,  . ​

Quizplus · Accepted Answer

To find horizontal tangents, we need to set the derivative of y with respect to x equal to 0 and solve for x. 
Taking the derivative of the given curve, we get:
dy/dx = (44x^3 - 60x^2 + 120x - 90)/(4x^2 - 4x + 5)^2
Setting the numerator equal to 0, we get:
44x^3 - 60x^2 + 120x - 90 = 0
Dividing by 2, we get:
22x^3 - 30x^2 + 60x - 45 = 0
Dividing by 3, we get:
22/3 x^3 - 10x^2 + 20x - 15/2 = 0
Using the rational root theorem, we can check that x = 1/2 is a root of this equation. We can use long division to factor out (2x - 1) and get:
22/3 x^3 - 10x^2 + 20x - 15/2 = (2x - 1)(11/3 x^2 - 4x + 15/2)
The quadratic factor has no real roots, so the only horizontal tangent occurs when 2x - 1 = 0, or x = 1/2. 
Plugging x = 1/2 into the original function, we get:
y = 11eac44f_ebf6_5086_bedf_2d02a177b8a1_TB7497_11
Therefore, there is one horizontal tangent at (1/2, 11eac44f_ebf6_5086_bedf_2d02a177b8a1_TB7497_11).
To find vertical tangents, we need to look for points where the derivative is undefined. In this case, the derivative is undefined when the denominator of the fraction is equal to zero, that is:
4x^2 - 4x + 5 = 0
Using the quadratic formula, we get:
x = (4 ± sqrt(4^2 - 4*4*5))/(2*4) = (1 ± i√6)/2
Since these are complex values, there are no points of vertical tangency.

Find All Points (If Any)of Horizontal and Vertical Tangency to the Curve