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

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

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

tangent of

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

y = - 6 x + 3

Solution steps

Find the slope of

f (x) = 5 x^{2} - 6 x + 3 : \frac{df ( x )}{d x} = 10 x - 6

Compute the slope of

f (x) = 5 x^{2} - 6 x + 3

(0, 3) : m = - 6

Find the line with slope m=

- 6

and passing through

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

y = - 6 x + 3

\int ((\frac{100 + x}{50}) - 17.33) d x

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

y^{^{'}} = 2 y - 3 y^{2}

\frac{d y}{d x} = x + 8

\frac{1}{2 ( s + 1 )}

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