What is the tangent of f(x)=5x^2-8x,\at x=1 ?

The tangent of f(x)=5x^2-8x,\at x=1 is y=2x-5

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tangent of f(x)=5x^2-8x,\at x=1

tangent of

f (x) = 5 x^{2} - 8 x, at x = 1

y = 2 x - 5

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = 5 x^{2} - 8 x : \frac{df ( x )}{d x} = 10 x - 8

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 = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 5

x \to 2 lim (a x + b)

\frac{2 x ^{2}}{x ^{2} - 1}

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{2}})

\frac{6 x - 1}{2 x + 9}

\int e^{- z^{2}} d z

What is the tangent of f(x)=5x^2-8x,\at x=1 ?
The tangent of f(x)=5x^2-8x,\at x=1 is y=2x-5