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=19x-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 = 19 x - 36

Solution steps

Find the tangent point:

(3, 21)

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 = 19

Find the line with slope m=

19

and passing through

(3, 21) : y = 19 x - 36

y = 19 x - 36

f (x) = - 4 x

\frac{\partial}{\partial y} (\frac{t}{2 y ^{4}})

\int x sin (5 x^{2}) cos (6 x^{2}) d x

\frac{d y}{d x} = x^{2} + y

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

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=19x-36