What is the tangent of f(x)=(4+5x)^2,(-3,121) ?

The tangent of f(x)=(4+5x)^2,(-3,121) is y=-110x-209

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tangent of f(x)=(4+5x)^2,(-3,121)

tangent of

f (x) = (4 + 5 x)^{2}, (- 3, 121)

y = - 110 x - 209

Solution steps

Find the slope of

f (x) = (4 + 5 x)^{2} : \frac{df ( x )}{d x} = 10 (4 + 5 x)

Compute the slope of

f (x) = (4 + 5 x)^{2}

(- 3, 121) : m = - 110

Find the line with slope m=

- 110

and passing through

(- 3, 121) : y = - 110 x - 209

y = - 110 x - 209

f (x) = 2 x, (18, 6)

\frac{d y}{d x} = \frac{e ^{y} + 1}{e ^{y}}

x = - 2, x = 1, y = 2 x^{2} + 7, y = 0

y^{^{'}} + 2 x y = 2 x e^{- x^{2}}

x \to 0 lim (\frac{x + 1}{x})

What is the tangent of f(x)=(4+5x)^2,(-3,121) ?
The tangent of f(x)=(4+5x)^2,(-3,121) is y=-110x-209