What is the tangent of f(x)=6x^2-10x+4,\at x=1 ?

The tangent of f(x)=6x^2-10x+4,\at x=1 is y=2x-2

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tangent of f(x)=6x^2-10x+4,\at x=1

tangent of

f (x) = 6 x^{2} - 10 x + 4, at x = 1

y = 2 x - 2

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 6 x^{2} - 10 x + 4 : \frac{df ( x )}{d x} = 12 x - 10

Compute the slope of

f (x) = 6 x^{2} - 10 x + 4

x = 1 : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 2

x \to 3 lim (ln (x))

\frac{d ^{2}}{d x ^{2}} (cos^{2} (x))

\int e^{2 y} - y cos (x y) d x

\frac{d}{d x} (4.9 x^{2})

\frac{\partial}{\partial x} (ln (\frac{x - 2}{3}))

What is the tangent of f(x)=6x^2-10x+4,\at x=1 ?
The tangent of f(x)=6x^2-10x+4,\at x=1 is y=2x-2