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Find the coordinates of ALL points on the curve where the line tangent to the curve is vertical. Curve: x^2-6x+y^3-12y=11, Derivative: dy/dx=(6-2x)/(3x^2-12)

User Timcbaoth
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8.2k points

1 Answer

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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)

User Arathorn
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8.3k points