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

The tangent of f(x)=2x^2-5x+1,\at x=2 is y=3x-7

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

tangent of

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

y = 3 x - 7

Solution steps

Find the tangent point:

(2, - 1)

Find the slope of

f (x) = 2 x^{2} - 5 x + 1 : \frac{df ( x )}{d x} = 4 x - 5

Compute the slope of

f (x) = 2 x^{2} - 5 x + 1

x = 2 : m = 3

Find the line with slope m=

3

and passing through

(2, - 1) : y = 3 x - 7

y = 3 x - 7

\int \frac{x ^{3}}{4 - x ^{2}} d x

(- 3, 1), (2, - 1)

\frac{d}{d x} (lo g_{10} (x^{2}))

\frac{d}{d x} ((x - \frac{1}{x}) (x^{2} - \frac{1}{x} - 4))

\frac{d}{d x} (0.6 x^{2})

What is the tangent of f(x)=2x^2-5x+1,\at x=2 ?
The tangent of f(x)=2x^2-5x+1,\at x=2 is y=3x-7