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

The tangent of f(x)=x^2+3,\at x=2 is y=4x-1

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

tangent of

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

y = 4 x - 1

Solution steps

Find the tangent point:

(2, 7)

Find the slope of

f (x) = x^{2} + 3 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 3

x = 2 : m = 4

Find the line with slope m=

4

and passing through

(2, 7) : y = 4 x - 1

y = 4 x - 1

f (x) = ln (x^{3} + 2 x^{2} + 2 x)

\int \frac{1}{x ( x ^{2} - 1 )} d x

\int \frac{( 2 x + 1 )}{x ^{2} - 2 x + 5} d x

x \to \frac{π}{4} lim (tan^{2} (x))

\int 9 tan (x) d x

What is the tangent of f(x)=x^2+3,\at x=2 ?
The tangent of f(x)=x^2+3,\at x=2 is y=4x-1