What is the tangent of y=x^2-3x+3,(4,7) ?

The tangent of y=x^2-3x+3,(4,7) is y=5x-13

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tangent of y=x^2-3x+3,(4,7)

tangent of

y = x^{2} - 3 x + 3, (4, 7)

y = 5 x - 13

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

(4, 7) : m = 5

Find the line with slope m=

5

and passing through

(4, 7) : y = 5 x - 13

y = 5 x - 13

\frac{d ^{2}}{d x ^{2}} (4 x + 1)

\int 2 x + 5 e^{x} d x

\frac{d}{d x} (8 x - 1)

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 1

x \to 9 lim (\frac{7 x}{x - 9})

What is the tangent of y=x^2-3x+3,(4,7) ?
The tangent of y=x^2-3x+3,(4,7) is y=5x-13