What is the tangent of y=5x^2-6x+3,(0,3) ?

The tangent of y=5x^2-6x+3,(0,3) is y=-6x+3

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tangent of y=5x^2-6x+3,(0,3)

tangent of

y = 5 x^{2} - 6 x + 3, (0, 3)

y = - 6 x + 3

Solution steps

Find the slope of

y = 5 x^{2} - 6 x + 3 : \frac{d y}{d x} = 10 x - 6

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 = - 6

Find the line with slope m=

- 6

and passing through

(0, 3) : y = - 6 x + 3

y = - 6 x + 3

x^{s i n (1)}

y = x^{3} + 1, y = x + 1

x \to 3 - lim (\frac{5}{x - 3})

\int 9 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

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

What is the tangent of y=5x^2-6x+3,(0,3) ?
The tangent of y=5x^2-6x+3,(0,3) is y=-6x+3