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

The tangent of y=x^2-5x+3,(2,-3) is y=-x-1

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

tangent of

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

y = - x - 1

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} - 5 x + 3

(2, - 3) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 1

\int x \cdot x^{2} - 1 d x

\int \frac{225 - x ^{2}}{x} d x

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

\frac{\partial}{\partial x} (sin (x + e^{y}))

8 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

What is the tangent of y=x^2-5x+3,(2,-3) ?
The tangent of y=x^2-5x+3,(2,-3) is y=-x-1