What is the tangent of f(x)=x^2+1,(0,1) ?

The tangent of f(x)=x^2+1,(0,1) is y=1

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tangent of f(x)=x^2+1,(0,1)

tangent of

f (x) = x^{2} + 1, (0, 1)

y = 1

Solution steps

Find the slope of

f (x) = x^{2} + 1 : \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 = 0

Find the line with slope m=

0

and passing through

(0, 1) : y = 1

y = 1

n = 5 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

t an g e n t y = x^{2} - 5

x \to - 1 lim (4 x^{2} - 2 x - 6)

d er i v a t i v e sin (2 π t)

\frac{d}{d y} (\frac{350}{x ^{2} + y ^{2}})

What is the tangent of f(x)=x^2+1,(0,1) ?
The tangent of f(x)=x^2+1,(0,1) is y=1