What is the tangent of f(x)=x^2+2,(-5,27) ?

The tangent of f(x)=x^2+2,(-5,27) is y=-10x-23

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tangent of f(x)=x^2+2,(-5,27)

tangent of

f (x) = x^{2} + 2, (- 5, 27)

y = - 10 x - 23

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 27) : y = - 10 x - 23

y = - 10 x - 23

x \to - \infty lim ((1)^{x})

\frac{\partial}{\partial x} (sin (- 5 x - 2 y + z))

x e^{- x}

\int_{0}^{\frac{π}{21}} cos (7 x) d x

\frac{141}{7 - 2}

What is the tangent of f(x)=x^2+2,(-5,27) ?
The tangent of f(x)=x^2+2,(-5,27) is y=-10x-23