What is the tangent of f(x)=x^2-5,\at x=3 ?

The tangent of f(x)=x^2-5,\at x=3 is y=6x-14

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tangent of f(x)=x^2-5,\at x=3

tangent of

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

y = 6 x - 14

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

f (x) = x^{2} - 5 : \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 = 6

Find the line with slope m=

6

and passing through

(3, 4) : y = 6 x - 14

y = 6 x - 14

\frac{\partial}{\partial y} (\frac{2}{2 x + 3 y})

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

- 3 cos (t)

\int (\frac{2 x}{x ^{2} + y ^{4}}) d x

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

What is the tangent of f(x)=x^2-5,\at x=3 ?
The tangent of f(x)=x^2-5,\at x=3 is y=6x-14