What is the tangent of f(x)=x^2+4,(4,20) ?

The tangent of f(x)=x^2+4,(4,20) is y=8x-12

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tangent of f(x)=x^2+4,(4,20)

tangent of

f (x) = x^{2} + 4, (4, 20)

y = 8 x - 12

Solution steps

Find the slope of

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

Find the line with slope m=

8

and passing through

(4, 20) : y = 8 x - 12

y = 8 x - 12

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

y^{^{'}} = y (\frac{1}{x} - y)

y = x^{2} - 4 x - 5, y = 2 x - 5

\int_{2.25}^{2.75} 2 d x

\frac{d}{d x} (arctan (3 x^{2} - x))

What is the tangent of f(x)=x^2+4,(4,20) ?
The tangent of f(x)=x^2+4,(4,20) is y=8x-12