What is the tangent of f(x)=-x^2-x+4,\at x=-1 ?

The tangent of f(x)=-x^2-x+4,\at x=-1 is y=x+5

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tangent of f(x)=-x^2-x+4,\at x=-1

tangent of

f (x) = - x^{2} - x + 4, at x = - 1

y = x + 5

Solution steps

Find the tangent point:

(- 1, 4)

Find the slope of

f (x) = - x^{2} - x + 4 : \frac{df ( x )}{d x} = - 2 x - 1

Compute the slope of

f (x) = - x^{2} - x + 4

x = - 1 : m = 1

Find the line with slope m=

1

and passing through

(- 1, 4) : y = x + 5

y = x + 5

y^{^{''}} + 5 y^{^{'}} + 6 y = - 10 t e^{2 t}

\frac{d}{d x} (\frac{x ^{6}}{2})

ln (x + x^{2} - 3)

\int_{0}^{1} 2 π (2 - y) (8 y^{2}) d y

\frac{\partial}{\partial x} (8 x^{6} y^{7} - 7 x^{5} y^{6})

What is the tangent of f(x)=-x^2-x+4,\at x=-1 ?
The tangent of f(x)=-x^2-x+4,\at x=-1 is y=x+5