What is the tangent of f(x)=5x^2-8x+3,\at x=1 ?

The tangent of f(x)=5x^2-8x+3,\at x=1 is y=2x-2

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tangent of f(x)=5x^2-8x+3,\at x=1

tangent of

f (x) = 5 x^{2} - 8 x + 3, at x = 1

y = 2 x - 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 5 x^{2} - 8 x + 3 : \frac{df ( x )}{d x} = 10 x - 8

Compute the slope of

f (x) = 5 x^{2} - 8 x + 3

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 0) : y = 2 x - 2

y = 2 x - 2

x \to 0 lim (tan (\frac{x}{2}))

x arcsin (x)

n = 1 \sum \infty n!^{2} (5^{n})

\int \frac{1}{2} x^{2} e^{- \frac{1}{2} x} d x

x^{2} e^{- x} + (ln (x))^{2}

What is the tangent of f(x)=5x^2-8x+3,\at x=1 ?
The tangent of f(x)=5x^2-8x+3,\at x=1 is y=2x-2