What is the tangent of f(x)=x^2+3,\at x=6 ?

The tangent of f(x)=x^2+3,\at x=6 is y=12x-33

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tangent of f(x)=x^2+3,\at x=6

tangent of

f (x) = x^{2} + 3, at x = 6

y = 12 x - 33

Solution steps

Find the tangent point:

(6, 39)

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 = 6 : m = 12

Find the line with slope m=

12

and passing through

(6, 39) : y = 12 x - 33

y = 12 x - 33

\int \frac{cosh ( x )}{x} d x

\frac{4 - 4 s}{4 s ^{2} + 4 s + 17}

f (x) = sec (x) + tan (x)

\int 8 x - 4 d x

\int 4 sin^{3} (6 x co) s^{3} 6 x d x

What is the tangent of f(x)=x^2+3,\at x=6 ?
The tangent of f(x)=x^2+3,\at x=6 is y=12x-33