What is the tangent of f(x)=-4x^2+7x-14,\at x=-2 ?

The tangent of f(x)=-4x^2+7x-14,\at x=-2 is y=23x+2

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tangent of f(x)=-4x^2+7x-14,\at x=-2

tangent of

f (x) = - 4 x^{2} + 7 x - 14, at x = - 2

y = 23 x + 2

Solution steps

Find the tangent point:

(- 2, - 44)

Find the slope of

f (x) = - 4 x^{2} + 7 x - 14 : \frac{df ( x )}{d x} = - 8 x + 7

Compute the slope of

f (x) = - 4 x^{2} + 7 x - 14

x = - 2 : m = 23

Find the line with slope m=

23

and passing through

(- 2, - 44) : y = 23 x + 2

y = 23 x + 2

\frac{\partial}{\partial x} (x^{c o s (x)})

e^{18 x}

x \to \infty lim (x^{(\frac{1}{x ^{2} - 1})})

\int (x^{2} - 1)^{4} d x

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

What is the tangent of f(x)=-4x^2+7x-14,\at x=-2 ?
The tangent of f(x)=-4x^2+7x-14,\at x=-2 is y=23x+2