What is the tangent of f(x)=3-5x+2x^2,(-9,210) ?

The tangent of f(x)=3-5x+2x^2,(-9,210) is y=-41x-159

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tangent of f(x)=3-5x+2x^2,(-9,210)

tangent of

f (x) = 3 - 5 x + 2 x^{2}, (- 9, 210)

y = - 41 x - 159

Solution steps

Find the slope of

f (x) = 3 - 5 x + 2 x^{2} : \frac{df ( x )}{d x} = 4 x - 5

Compute the slope of

f (x) = 3 - 5 x + 2 x^{2}

(- 9, 210) : m = - 41

Find the line with slope m=

- 41

and passing through

(- 9, 210) : y = - 41 x - 159

y = - 41 x - 159

(x + y)^{2} d x + (2 x y + x^{2} - 9) d y = 0

x \to \infty lim (\frac{x}{2 x})

\frac{\partial}{\partial x} (\frac{x}{2 + y})

f (t) = cos (ln (t))

x y^{^{'}} - y = \frac{5}{2} x ln (x)

What is the tangent of f(x)=3-5x+2x^2,(-9,210) ?
The tangent of f(x)=3-5x+2x^2,(-9,210) is y=-41x-159