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

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

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

tangent of

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

y = - 2 x + 2

Solution steps

Find the slope of

y = x^{4} - 3 x^{2} + 2 : \frac{d y}{d x} = 4 x^{3} - 6 x

Compute the slope of

y = x^{4} - 3 x^{2} + 2

(1, 0) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

x \to 2 - lim (\frac{∣ x - x ∣}{x - 2})

\frac{d}{d x} (tan (3 x) s c^{3} x)

\int \frac{1}{x 1 - 4 x ^{2}} d x

f (x) = \frac{x ^{4}}{4} - 2 x^{2} - 4

x^{2} - 2 x + 4

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