What is the tangent of f(x)=-4x^2-3x+2,\at x=-1 ?

The tangent of f(x)=-4x^2-3x+2,\at x=-1 is y=5x+6

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tangent of f(x)=-4x^2-3x+2,\at x=-1

tangent of

f (x) = - 4 x^{2} - 3 x + 2, at x = - 1

y = 5 x + 6

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = - 4 x^{2} - 3 x + 2 : \frac{df ( x )}{d x} = - 8 x - 3

Compute the slope of

f (x) = - 4 x^{2} - 3 x + 2

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

(- 1, 1) : y = 5 x + 6

y = 5 x + 6

\int_{1}^{3} x^{2} \frac{2.25}{x ^{3}} d x

\int (3 x + 1)^{5} d x

\frac{d}{d θ} (\frac{5}{9 sin ( θ ) + 7 cos ( θ )})

\frac{d}{d x} (3 (e^{x} + x e^{x}))

x^{2} - 4 x, 3 x, 0, 7

What is the tangent of f(x)=-4x^2-3x+2,\at x=-1 ?
The tangent of f(x)=-4x^2-3x+2,\at x=-1 is y=5x+6