What is the tangent of f(x)=x^2+3,\at x=1 ?

The tangent of f(x)=x^2+3,\at x=1 is y=2x+2

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tangent of f(x)=x^2+3,\at x=1

tangent of

f (x) = x^{2} + 3, at x = 1

y = 2 x + 2

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = x^{2} + 3 : \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 = 2

Find the line with slope m=

2

and passing through

(1, 4) : y = 2 x + 2

y = 2 x + 2

x \to - 10 - lim (\frac{x + 9}{x + 10})

\frac{\partial}{\partial x} (2 e^{- y} x)

\int \frac{sin ( 2 x )}{37 + cos ^{2} ( x )} d x

\int x \cdot lo g_{10} (x) d x

m \to 4 lim (\frac{3 m ^{2} - 8 m - 16}{2 m ^{2} - 9 m + 4})

What is the tangent of f(x)=x^2+3,\at x=1 ?
The tangent of f(x)=x^2+3,\at x=1 is y=2x+2