What is the tangent of f(x)=4x^2+2x+5,\at x=3 ?

The tangent of f(x)=4x^2+2x+5,\at x=3 is y=26x-31

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tangent of f(x)=4x^2+2x+5,\at x=3

tangent of

f (x) = 4 x^{2} + 2 x + 5, at x = 3

y = 26 x - 31

Solution steps

Find the tangent point:

(3, 47)

Find the slope of

f (x) = 4 x^{2} + 2 x + 5 : \frac{df ( x )}{d x} = 8 x + 2

Compute the slope of

f (x) = 4 x^{2} + 2 x + 5

x = 3 : m = 26

Find the line with slope m=

26

and passing through

(3, 47) : y = 26 x - 31

y = 26 x - 31

t (\frac{d y}{d t}) + 33 y = \frac{1}{t}

\frac{d}{d x} (- 4 x^{2} + x + 5)

x \to π - lim (ln (sin (x)))

f (x) = 9 x e^{x}

\frac{\partial}{\partial x} (6 y x^{5} - 12 y^{2} x^{2})

What is the tangent of f(x)=4x^2+2x+5,\at x=3 ?
The tangent of f(x)=4x^2+2x+5,\at x=3 is y=26x-31