What is the tangent of y=x^2-3x+3,(0,3) ?

The tangent of y=x^2-3x+3,(0,3) is y=-3x+3

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tangent of y=x^2-3x+3,(0,3)

tangent of

y = x^{2} - 3 x + 3, (0, 3)

y = - 3 x + 3

Solution steps

Find the slope of

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

Compute the slope of

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

(0, 3) : m = - 3

Find the line with slope m=

- 3

and passing through

(0, 3) : y = - 3 x + 3

y = - 3 x + 3

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

f (x) = \frac{A}{x ^{7}} + B e^{7}

\frac{d}{d x} (cos (2))

\int_{1}^{3} (x^{2} + 1)^{3} d x

\frac{d ^{2}}{d x ^{2}} (8 x cos (x^{2}))

What is the tangent of y=x^2-3x+3,(0,3) ?
The tangent of y=x^2-3x+3,(0,3) is y=-3x+3