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

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

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

tangent of

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

y = - 2 x + 2

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} + 3

(- 1, 4) : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 2

\int_{- 4}^{4} ∣ x ∣ d x

x \to - 3 lim (\frac{4 x - 5}{3 x + 9})

x \to 1 lim ([3 x + 1])

\int (1 + 6 x)^{4} (6) d x

\int e^{4 \cdot l n (x)} d x

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