What is the tangent of y=-1/4 x^2,(-2,-1) ?

The tangent of y=-1/4 x^2,(-2,-1) is y=x+1

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tangent of y=-1/4 x^2,(-2,-1)

tangent of

y = - \frac{1}{4} x^{2}, (- 2, - 1)

y = x + 1

Solution steps

Find the slope of

y = - \frac{1}{4} x^{2} : \frac{d y}{d x} = - \frac{x}{2}

Compute the slope of

y = - \frac{1}{4} x^{2}

(- 2, - 1) : m = 1

Find the line with slope m=

1

and passing through

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

y = x + 1

(1 + 2 x)^{2}

y^{^{''}} - 8 y^{^{'}} + 41 y = 0

\frac{d}{d x} ((x^{2} - x + 1)^{- 7})

\int \frac{x - 1}{x ( x ^{2} - 5 x + 6 )} d x

\int (3 - 5 x)^{2} d x

What is the tangent of y=-1/4 x^2,(-2,-1) ?
The tangent of y=-1/4 x^2,(-2,-1) is y=x+1