What is the tangent of x^2-3x+6,\at x=1 ?

The tangent of x^2-3x+6,\at x=1 is y=-x+5

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tangent of x^2-3x+6,\at x=1

tangent of

x^{2} - 3 x + 6, at x = 1

y = - x + 5

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

y = x^{2} - 3 x + 6 : \frac{d y}{d x} = 2 x - 3

Compute the slope of

y = x^{2} - 3 x + 6

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 4) : y = - x + 5

y = - x + 5

\frac{d}{d x} (7 x^{2} - \frac{3}{x} + 2 x)

\frac{d}{d x} (\frac{x - 3}{x + 3})

x^{2} + 7

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

\frac{d x}{d t} + t^{7} x^{9} + \frac{x}{t} = 0

What is the tangent of x^2-3x+6,\at x=1 ?
The tangent of x^2-3x+6,\at x=1 is y=-x+5