# What is the equation of the tangent to the curve # x^3 +y^3 +3x-6y=0 # at the coordinate #(1,2)#?

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Gradient is

Differentiating the equation with respect to

#3x^2 +3y^2dy/dx +3-6dy/dx=0#

#(3y^2-6)dy/dx =-3x^2-3#

#dy/dx =-(3x^2+3)/(3y^2-6)#

#dy/dx |_("("1,2")")=-(3xx 1^2+3)/(3xx2^2-6)#

#=>=-6/(6)#

#=>=-1#

.-.-.-.-.-.-.-.-

Possible if the equation is

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So

If we modify the equation and use

It has been confirmed in the comments that

The following will show that in fact it cannot be:

We have:

# x^3 +y^3 +3x-6y+9=0 #

With

# LHS = 1+8+3-12+9 = 9 != 0 #

So

Here is a graph to confirm this visually:

If as Jim H suggests that the correct equation is:

# x^3 +y^3 +3x-6y=0 # ..... [A]Then now with

#x=1; y=2# we have:

# LHS = 1+8+3-12 = 0 # So

#(1,2)# does satisfy the equation of the modified equation:Differentiating [A] implicitly we have:

# 3x^2 +3y^2dy/dx +3-6dy/dx=0 # We could rearrange this equation and form an explicit expression for

#dy/dx# , but this isn't strictly necessary, we just need the value at#(1,2)#

with#x=1; y=2# we have:

# 3 +12dy/dx +3-6dy/dx=0 #

# :. 6 +6dy/dx=0 => dy/dx = -1# And this is the gradient of the tangent

We can go on and determine the equation of the tangent

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is

#−1# ).Using the point/slope form

#y-y_1=m(x-x_1)# , the tangent equation we seek:

# y - 2 = (-1)(x-1) #

# :. y - 2 = -x+1#

# :. y = -x+3 # The graphs of the equation are:

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To find the equation of the tangent to the curve at the coordinate (1,2), we need to find the slope of the tangent line at that point.

First, we differentiate the equation of the curve implicitly with respect to x:

3x^2 + 3y^2(dy/dx) + 3 - 6(dy/dx) = 0

Next, we substitute the coordinates (1,2) into the equation:

3(1)^2 + 3(2)^2(dy/dx) + 3 - 6(dy/dx) = 0

Simplifying this equation gives:

3 + 12(dy/dx) + 3 - 6(dy/dx) = 0

Combining like terms:

6(dy/dx) + 6 = 0

Solving for dy/dx:

dy/dx = -1

The slope of the tangent line at the point (1,2) is -1.

Using the point-slope form of a line, we can write the equation of the tangent line:

y - 2 = -1(x - 1)

Simplifying this equation gives:

y - 2 = -x + 1

Rearranging the equation:

x + y = 3

Therefore, the equation of the tangent to the curve at the coordinate (1,2) is x + y = 3.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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