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

The tangent of y=x^2-3x,(1,-2) is y=-x-1

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

tangent of

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

y = - x - 1

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 3 x

(1, - 2) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 1

y^{^{''}} + 2 y^{^{'}} + 5 y = 3 cos (x)

t y^{^{'}} + (t + 1) y = t, y (ln (2)) = 1

\int (6 sin (2 x) + sec^{2} (x)) d x

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{4 + x})

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