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

The tangent of f(x)=x^2-2x+2,\at x=4 is y=6x-14

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

tangent of

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

y = 6 x - 14

Solution steps

Find the tangent point:

(4, 10)

Find the slope of

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

Compute the slope of

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

x = 4 : m = 6

Find the line with slope m=

6

and passing through

(4, 10) : y = 6 x - 14

y = 6 x - 14

\frac{d}{d x} (ln (\frac{5 x ^{2} + 3}{x ^{2}}))

3 x y^{^{'}} + 7 y = 0, y (1) = 8

\int 2 (2 x + 1) 6 d x

y^{^{''}} + 8 y^{^{'}} - 9 y = 0

(z^{2} + e^{z}) z

What is the tangent of f(x)=x^2-2x+2,\at x=4 ?
The tangent of f(x)=x^2-2x+2,\at x=4 is y=6x-14