What is the tangent of f(x)=x^2-2x+4,\at x=1 ?

The tangent of f(x)=x^2-2x+4,\at x=1 is y=3

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tangent of f(x)=x^2-2x+4,\at x=1

tangent of

f (x) = x^{2} - 2 x + 4, at x = 1

y = 3

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

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

(1, 3) : y = 3

y = 3

\frac{d}{d x} (- x^{2} y)

3 - 6 x

y^{^{''}} - 2 y^{^{'}} + y = 6 e^{x}

\int_{0}^{4} 5 x - x^{2} - x d x

\frac{\partial}{\partial x} (ln (5 + x^{2} y^{2}))

What is the tangent of f(x)=x^2-2x+4,\at x=1 ?
The tangent of f(x)=x^2-2x+4,\at x=1 is y=3