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

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

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

tangent of

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

y = (2 a + 6) x - a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, a^{2} + 6 a)

Find the slope of

f (x) = x^{2} + 6 x : \frac{df ( x )}{d x} = 2 x + 6

Compute the slope of

f (x) = x^{2} + 6 x

x = a : m = 2 a + 6

Find the line with slope m=

2 a + 6

and passing through

(a, a^{2} + 6 a) : y = (2 a + 6) x - a^{2}

y = (2 a + 6) x - a^{2}

x \to 2 lim (3 x - 2)

\frac{x ^{2}}{8 + x}

\int_{0}^{1} 23 cos (x^{2}) d x

\frac{d}{d x} ((tan (x))^{\frac{1}{x}})

\int \frac{1}{x ^{2} + 5 x - 6} d x

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