What is the tangent of f(x)=2+4x^2,(0,2) ?

The tangent of f(x)=2+4x^2,(0,2) is y=2

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tangent of f(x)=2+4x^2,(0,2)

tangent of

f (x) = 2 + 4 x^{2}, (0, 2)

y = 2

Solution steps

Find the slope of

f (x) = 2 + 4 x^{2} : \frac{df ( x )}{d x} = 8 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 = 0

Find the line with slope m=

0

and passing through

(0, 2) : y = 2

y = 2

y = \frac{8}{sin ( x ) + cos ( x )}, (0, 8)

\frac{\partial}{\partial x} (\frac{3 x - y}{x + 2 y})

\int \frac{cos ( x )}{sin ( x ) + sin ^{2} ( x )} d x

y^{^{''}} + 34 y = 0

d x + d y e^{6 x} = 0

What is the tangent of f(x)=2+4x^2,(0,2) ?
The tangent of f(x)=2+4x^2,(0,2) is y=2