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

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

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

tangent of

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

y = 5 x - 7

Solution steps

Find the slope of

f (x) = 2 x^{2} - 3 x + 1 : \frac{df ( x )}{d x} = 4 x - 3

Compute the slope of

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

(2, 3) : m = 5

Find the line with slope m=

5

and passing through

(2, 3) : y = 5 x - 7

y = 5 x - 7

\frac{d}{d x} (\frac{x + 1}{x ^{2}})

\frac{\partial}{\partial z} ((x y)^{z})

f (x) = cos (π x)

\frac{d}{d x} (- 4 x^{5})

\frac{d}{d x} (\frac{x - 4}{7 x ^{2} e ^{x}})

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