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

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

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

tangent of

f (x) = x^{2} - 4 x + 3, at x = 1

y = - 2 x + 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{2} - 4 x + 3 : \frac{df ( x )}{d x} = 2 x - 4

Compute the slope of

f (x) = x^{2} - 4 x + 3

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 0) : y = - 2 x + 2

y = - 2 x + 2

\int \frac{15 arctan ( x )}{x ^{2}} d x

\int_{x^{2}}^{1} ln (2 t^{2} + 1) d t

n = 5 \sum \infty e^{3 - 2 n}

\int 3 t^{2} sin (2 t^{3} + 1) d t

\int \frac{2 x}{x} d x

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