**Q: If x ^{2} +xy + y^{2} = 3, what is the equation of the tangent line to this graph when y = 1.**

A: To find the equation of a tangent line, we know our final answer will be in **y=mx+b** format. To get this, we will need to find a point and a slope.

**Step 1**: Establish our point of interest. When y = 1, we need to know what x equals. Plug 1 into the equation for y to get:

x^{2} +x*(1) + (1)^{2} = 3

x^{2} +x – 2 = 0

(x + 2)(x – 1) = 0

x = -2, 1.

So, we have two points of interest: (x, y) = (-2, 1) and (1, 1). We will have 2 different tangent lines when y = 1.

**Step 2**: Finding the slope(s): To find the slope at one point, we need a derivative!

Take the derivative of this equation implicitly as follows (let y’ be the same as dy/dx):

d/dx(x^{2} +xy + y^{2} = 3) –> **2x + y + xy’ + 2yy’ = 0**

Note: Implicit differentiation above required a product rule and a chain rule. We are taking the derivative of each piece with respect to the variable x!

Now, solve for y’ (since y’ is the slope):

2x + y + xy’ + 2yy’ = 0

xy’ + 2yy’ = -2x – y

y'(x + 2y) = -2x – y

y’ = (-2x – y) / (x + 2y)

Now, plug in each point separately to find the slope at the given point:

Point One, (-2, 1), yields:

y’ = undefined (dividing by zero). The slope is undefined at this point, so there is no tangent line.

Point Two, (1, 1), yields:

y’ = -1.

So, the slope at the point (1, 1) is -1

**Step Three: **The equation of the tangent line:

y = mx + b

Plug in our point and slope to solve for b!

1 = -1*1 + b

1 = -1 + b

2 = b.

The equation of the tangent line at the point (1, 1) to the given graph is **y = -x+2**!