Explanation:
your made a typo in your problem definition. the real derivative is
dy/dx = (6 - 2x)/(3y² - 12)
a vertical line has an undefined (or often declared as infinite) slope, as all the points on such a line have the same x-value.
the slope of a line is the ratio
y coordinate difference / x coordinate difference
when going from one point on the line to another.
for a vertical line the x coordinate difference is always 0, and therefore the slope is in the form y/0. which is undefined (or infinite).
bringing the 2 points in the line closer and closer together with the limit in infinity to become the same, gives us the slope of the tangent at this point.
dy/dx is therefore the function of all the slopes along the original curve.
we need to find the points (x-values), where dy/dx is undefined (or infinite).
and we know, this happens for a fraction, when the denominator (bottom part) turns 0.
therefore, we need to find the x-values, so that
3y² - 12 = 0
y² - 4 = 0
y² = 4
y = ±sqrt(4) = ±2
the curve has vertical tangents at
y = -2 and y = 2
for the x-values we need to use these y-values in the curve equation and solve for x :
x² - 6x + (-2)³ - 12×-2 = 11
x² - 6x - 8 + 24 = 11
x² - 6x + 16 = 11
x² - 6x + 5 = 0
the solution for a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -6
c = 5
x = (6 ± sqrt((-6)² - 4×1×5))/(2×1) =
= (6 ± sqrt(36 - 20))/2 = (6 ± sqrt(16))/2 =
= (6 ± 4)/2 = 3 ± 2
so, for y = -2 we get
x = 3+2 = 5 and x = 3-2 = 1
now for y = 2 :
x² - 6x + 2³ - 12×2 = 11
x² - 6x + 8 - 24 = 11
x² - 6x - 16 = 11
x² - 6x - 27 = 0
using the same formula and approach as above we get
x = (6 ± sqrt((-6)² - 4×1×-27))/(2×1) =
= (6 ± sqrt(36 + 108))/2 = (6 ± sqrt(144))/2 =
= (6 ± 12)/2 = 3 ± 6
so, for y = 2 we get
x = 3+6 = 9 and x = 3-6 = -3
the points with vertical tangents are therefore
(5, -2), (1, -2), (9, 2), (-3, 2)