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

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

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tangent of f(x)=-x^2+x,(-1,-2)

tangent of

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

y = 3 x + 1

Solution steps

Find the slope of

f (x) = - x^{2} + x : \frac{df ( x )}{d x} = - 2 x + 1

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 = 3

Find the line with slope m=

3

and passing through

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

y = 3 x + 1

\int (x - 4)^{2} d x

y^{^{''}} + 4 y^{^{'}} = 17 cos (t), y (0) = - 1, y^{^{'}} (0) = 0

\frac{d}{d x} (- \frac{16}{( 1 + 4 x ) ^{2}})

\int_{1}^{2} 21 t^{2} d t

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

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