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

The tangent of f(x)=4x^2-5x,\at x=-3 is y=-29x-36

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tangent of f(x)=4x^2-5x,\at x=-3

tangent of

f (x) = 4 x^{2} - 5 x, at x = - 3

y = - 29 x - 36

Solution steps

Find the tangent point:

(- 3, 51)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 29

Find the line with slope m=

- 29

and passing through

(- 3, 51) : y = - 29 x - 36

y = - 29 x - 36

\frac{\partial}{\partial x} (x^{2} ln (y - 1))

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

\int ln (t) \cdot t^{2} d t

x \to + (- \infty) + lim (x^{5})

\frac{d y}{d x} = \frac{5 x}{2 y}

What is the tangent of f(x)=4x^2-5x,\at x=-3 ?
The tangent of f(x)=4x^2-5x,\at x=-3 is y=-29x-36