What is the tangent of 0=x^2-xy+2y-2,\at x=-2 ?

The tangent of 0=x^2-xy+2y-2,\at x=-2 is y= 7/8 x+5/4

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tangent of 0=x^2-xy+2y-2,\at x=-2

tangent of

0 = x^{2} - x y + 2 y - 2, at x = - 2

y = \frac{7}{8} x + \frac{5}{4}

Solution steps

Find the tangent point:

(- 2, - \frac{1}{2})

Find the slope of

0 = x^{2} - x y + 2 y - 2 : \frac{d y}{d x} = \frac{x ^{2} - 4 x + 2}{( - x + 2 ) ^{2}}

Compute the slope of

0 = x^{2} - x y + 2 y - 2

x = - 2 : m = \frac{7}{8}

Find the line with slope m=

\frac{7}{8}

and passing through

(- 2, - \frac{1}{2}) : y = \frac{7}{8} x + \frac{5}{4}

y = \frac{7}{8} x + \frac{5}{4}

\frac{\partial}{\partial x} (x^{3} \cdot y^{2} + 4 x^{2} - 3 x y)

x \to \infty lim (x - {x})

\int 2 yz d y

x \to 0 lim (\frac{( 2 + x ) ^{2} - 8}{x})

y = x + \frac{4}{x}, (- 4, - 5)

What is the tangent of 0=x^2-xy+2y-2,\at x=-2 ?
The tangent of 0=x^2-xy+2y-2,\at x=-2 is y= 7/8 x+5/4