What is the tangent of y=2x^2-3x+1,(3,10) ?

The tangent of y=2x^2-3x+1,(3,10) is y=9x-17

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tangent of y=2x^2-3x+1,(3,10)

tangent of

y = 2 x^{2} - 3 x + 1, (3, 10)

y = 9 x - 17

Solution steps

Find the slope of

y = 2 x^{2} - 3 x + 1 : \frac{d y}{d x} = 4 x - 3

Compute the slope of

y = 2 x^{2} - 3 x + 1

(3, 10) : m = 9

Find the line with slope m=

9

and passing through

(3, 10) : y = 9 x - 17

y = 9 x - 17

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

\frac{d}{d x} (x^{0.4})

e^{2 t} cos (5) (t)

e^{2 x}, e^{6 x}, x = 1

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

What is the tangent of y=2x^2-3x+1,(3,10) ?
The tangent of y=2x^2-3x+1,(3,10) is y=9x-17