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

The tangent of y=x^2-3x,(3,0) is y=3x-9

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

tangent of

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

y = 3 x - 9

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3

Find the line with slope m=

3

and passing through

(3, 0) : y = 3 x - 9

y = 3 x - 9

y^{^{'}} - 3 y = 4 e^{5 t}, y (0) = 5

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

\int \frac{1}{x ( x - 8 ) ^{2}} d x

\frac{d}{d x} (π x^{2} y)

\int_{0}^{1} ln (9 x) d x