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

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

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

tangent of

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

y = 3 x - 5

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

f (x) = 2 x^{2} - 5 x + 3 : \frac{df ( x )}{d x} = 4 x - 5

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 = 3

Find the line with slope m=

3

and passing through

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

y = 3 x - 5

\frac{d}{d t} (\frac{8 t}{8 + t})

\int \frac{1}{t ^{2} - 8 t + 20} d t

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

y^{^{''}} + 7 y^{^{'}} - 8 y = 0

\frac{d y}{d x} = 6 x^{5} e^{- y}

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