What is the tangent of f(x)=12x-1.86x^2,\at x=a ?

The tangent of f(x)=12x-1.86x^2,\at x=a is y=(12-3.72a)x+1.86a^2

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tangent of f(x)=12x-1.86x^2,\at x=a

tangent of

f (x) = 12 x - 1.86 x^{2}, at x = a

y = (12 - 3.72 a) x + 1.86 a^{2}

Solution steps

Find the tangent point:

(a, 12 a - 1.86 a^{2})

Find the slope of

f (x) = 12 x - 1.86 x^{2} : \frac{df ( x )}{d x} = 12 - 3.72 x

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 = 12 - 3.72 a

Find the line with slope m=

12 - 3.72 a

and passing through

(a, 12 a - 1.86 a^{2}) : y = (12 - 3.72 a) x + 1.86 a^{2}

y = (12 - 3.72 a) x + 1.86 a^{2}

y = (3 x + 1) 8 x - 1

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

x \to \infty lim (5 x + 2)

\frac{d}{d x} y = \frac{1}{( 8 x - 5 ) ^{5}}

y^{^{'}} - 2 y = 4 x

What is the tangent of f(x)=12x-1.86x^2,\at x=a ?
The tangent of f(x)=12x-1.86x^2,\at x=a is y=(12-3.72a)x+1.86a^2