What is the tangent of f(x)=x^2-3x,\at x=1 ?

The tangent of f(x)=x^2-3x,\at x=1 is y=-x-1

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tangent of f(x)=x^2-3x,\at x=1

tangent of

f (x) = x^{2} - 3 x, at x = 1

y = - x - 1

Solution steps

Find the tangent point:

(1, - 2)

Find the slope of

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

Compute the slope of

f (x) = x^{2} - 3 x

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x - 1

\int (12 - 3 t^{2}) d t

\frac{2 s}{( s + 3 ) ^{2} + 7}

x \to - 2 lim ((x + 1) \cdot (x + 1))

\int 7 x + 1 d x

What is the tangent of f(x)=x^2-3x,\at x=1 ?
The tangent of f(x)=x^2-3x,\at x=1 is y=-x-1