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

The tangent of f(x)=4x^2-5x,\at x=-1 is y=-13x-4

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

tangent of

f (x) = 4 x^{2} - 5 x, at x = - 1

y = - 13 x - 4

Solution steps

Find the tangent point:

(- 1, 9)

Find the slope of

f (x) = 4 x^{2} - 5 x : \frac{df ( x )}{d x} = 8 x - 5

Compute the slope of

f (x) = 4 x^{2} - 5 x

x = - 1 : m = - 13

Find the line with slope m=

- 13

and passing through

(- 1, 9) : y = - 13 x - 4

y = - 13 x - 4

\int \frac{x ^{3} \cdot ( 1 - x ) ^{2}}{2} d x

\frac{d}{d t} (2 e^{t} sin (t))

\int (sec (x) tan^{5} (x))^{2} d x

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

What is the tangent of f(x)=4x^2-5x,\at x=-1 ?
The tangent of f(x)=4x^2-5x,\at x=-1 is y=-13x-4