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

The tangent of f(x)=2x^2-5x,\at x=3 is y=7x-18

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

tangent of

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

y = 7 x - 18

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

f (x) = 2 x^{2} - 5 x : \frac{df ( x )}{d x} = 4 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 = 7

Find the line with slope m=

7

and passing through

(3, 3) : y = 7 x - 18

y = 7 x - 18

\int_{0}^{3} 9 x - x^{3} d x

\int (tan (x))^{3} d x

\int 5 sin^{2} (x co) s^{3} x d x

x \to 5 - lim (\frac{x - 5}{∣ x - 5 ∣})

\int sinh (x) d x

What is the tangent of f(x)=2x^2-5x,\at x=3 ?
The tangent of f(x)=2x^2-5x,\at x=3 is y=7x-18