What is the tangent of f(x)=x^2-4x+3,(1,0) ?

The tangent of f(x)=x^2-4x+3,(1,0) is y=-2x+2

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tangent of f(x)=x^2-4x+3,(1,0)

tangent of

f (x) = x^{2} - 4 x + 3, (1, 0)

y = - 2 x + 2

Solution steps

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

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

\int_{0}^{π} sin (x^{2}) d x

\int cot^{3} (x) d x

f (x) = x + 8, (8, 4)

\int \frac{2 x}{9 - 4 x ^{2}} d x

\frac{12}{0.5 s ^{2} + 4 s + 5}

What is the tangent of f(x)=x^2-4x+3,(1,0) ?
The tangent of f(x)=x^2-4x+3,(1,0) is y=-2x+2