What is the tangent of f(x)=x^2+3,(3,12) ?

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

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tangent of f(x)=x^2+3,(3,12)

tangent of

f (x) = x^{2} + 3, (3, 12)

y = 6 x - 6

Solution steps

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

(3, 12) : m = 6

Find the line with slope m=

6

and passing through

(3, 12) : y = 6 x - 6

y = 6 x - 6

\int (3 x^{2} tan (y) - \frac{2 y ^{3}}{x ^{3}}) d x

f (x) = 3 csc (4 x)

(y^{4} - x^{2}) d y + x y d x = 0

x^{2} - 5 x

x y^{^{'}} + 4 y = 9 x^{5} + 2 x^{3}

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