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

Compute the slope of

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

(5, 27) : m = 10

Find the line with slope m=

10

and passing through

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

y = 10 x - 23

10 sin (x) cos (x)

\frac{d y}{d x}, y = 3 x^{2}

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

ln (x e^{- 8 x})

y = x^{4} - 2 x^{3}, y = 0

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