What is the tangent of x^2+3x+9,\at x=4 ?

The tangent of x^2+3x+9,\at x=4 is y=11x-7

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tangent of x^2+3x+9,\at x=4

tangent of

x^{2} + 3 x + 9, at x = 4

y = 11 x - 7

Solution steps

Find the tangent point:

(4, 37)

Find the slope of

y = x^{2} + 3 x + 9 : \frac{d y}{d x} = 2 x + 3

Compute the slope of

y = x^{2} + 3 x + 9

x = 4 : m = 11

Find the line with slope m=

11

and passing through

(4, 37) : y = 11 x - 7

y = 11 x - 7

\frac{d}{d x} ((x + \frac{1}{x})^{2} - \frac{5}{3 x})

\frac{\partial}{\partial x} (x e^{3 y} sin (5 z))

n = 1 \sum \infty (\frac{5}{6})^{n}

y = e^{k t a n (7 x)}

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

What is the tangent of x^2+3x+9,\at x=4 ?
The tangent of x^2+3x+9,\at x=4 is y=11x-7