Find the equation of the tangent line to the curve x^2 + xy + y^2 = 3 at the point (1,1)

y = -x + 2

Step-by-step explanation:

We first need to find the derivative of the equation x² + xy + y² = 3. this is done with implicit differentiation

2x + y + xy' + 2yy' = 0

Get terms with y' to one side, and the other terms to the other side of the equals sign...

xy' + 2yy' = -2x - y

Factor out y'...

y'(x + 2y) = -2x - y

Divide both sides by x + 2y

y' = (-2x - y)/(x + 2y)

This is the formula for the slope of the lines tangent to the curve of the original function.

Plug in the given point (1, 1) to find the slope of this

y' = (-2[1] - 1)/(1 + 2([1])

y' = -3/3

y' = -1

To find the equation of the line:

The general form for a linear equation in slope-intercept form is  

y = mx + b where m is the slope and b is the y-intercept

Use what we know about our line to solve the rest. By using one of the given points, we have 3 of the 4 variables in the above equation. Pick either point, it doesn't matter which one. We'll use (1, 1), which is (x, y), and we know that  

m = -1

The equation becomes...

1 = -1(1) + b (now solve for b)

1 = -1 + b

2 = b  

Plug that value into the general form...

y = -x + 2

See attached photo for the graphs of the original equation, and the graph of the tangent line at (1, 1)

what is the original number? you forgot to add it.

mx+89b(-98)nb

step-by-step explanation: