What is the tangent of f(x)=x^2+3,(4,19) ?

The tangent of f(x)=x^2+3,(4,19) is y=8x-13

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tangent of f(x)=x^2+3,(4,19)

tangent of

f (x) = x^{2} + 3, (4, 19)

y = 8 x - 13

Solution steps

Find the slope of

f (x) = x^{2} + 3 : \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 = 8

Find the line with slope m=

8

and passing through

(4, 19) : y = 8 x - 13

y = 8 x - 13

e^{2 x} sin (2 y) d x + (2 - e^{2 x}) sec (2 y) d y = 0

\frac{\partial}{\partial x} (\frac{ln ( 2 x ^{2} + 5 )}{y ^{2}})

\frac{\partial}{\partial y} (\frac{y ^{2}}{2})

\frac{d}{d x} (\frac{x ^{5}}{1 - x ^{4}})

\frac{d}{d x} ((x^{2} + 2)^{3})

What is the tangent of f(x)=x^2+3,(4,19) ?
The tangent of f(x)=x^2+3,(4,19) is y=8x-13