What is the tangent of f(x)=3x^2+2x,\at x=1 ?

The tangent of f(x)=3x^2+2x,\at x=1 is y=8x-3

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tangent of f(x)=3x^2+2x,\at x=1

tangent of

f (x) = 3 x^{2} + 2 x, at x = 1

y = 8 x - 3

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = 3 x^{2} + 2 x : \frac{df ( x )}{d x} = 6 x + 2

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8

Find the line with slope m=

8

and passing through

(1, 5) : y = 8 x - 3

y = 8 x - 3

n = 1 \sum \infty (- 0.3)^{n}

x \to 0 lim (\frac{1}{3} x)

f (x) = 3 x - cos (x), at x = 0

\int (9 x + 5)^{2} d x

f (x) = x^{- 5}

What is the tangent of f(x)=3x^2+2x,\at x=1 ?
The tangent of f(x)=3x^2+2x,\at x=1 is y=8x-3