What is the tangent of f(x)=60+20x-4.5x^2,\at x=2 ?

The tangent of f(x)=60+20x-4.5x^2,\at x=2 is y=2x+78

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tangent of f(x)=60+20x-4.5x^2,\at x=2

tangent of

f (x) = 60 + 20 x - 4.5 x^{2}, at x = 2

y = 2 x + 78

Solution steps

Find the tangent point:

(2, 82)

Find the slope of

f (x) = 60 + 20 x - 4.5 x^{2} : \frac{df ( x )}{d x} = - 9 x + 20

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 = 2

Find the line with slope m=

2

and passing through

(2, 82) : y = 2 x + 78

y = 2 x + 78

- x^{2} - x + 9, - x + 7, 0

\int sin^{3} (3 x) cos^{4} (3 x) d x

\frac{\partial}{\partial x} ((1 - x) (1 - y))

x \to 4 + lim (\frac{x - 4}{x ^{2} - 1})

f (x) = 3 x^{6} + 9 x^{2} - 7

What is the tangent of f(x)=60+20x-4.5x^2,\at x=2 ?
The tangent of f(x)=60+20x-4.5x^2,\at x=2 is y=2x+78