What is the tangent of y=x^2+3,(4,19) ?

The tangent of y=x^2+3,(4,19) is y=8x-13

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tangent of y=x^2+3,(4,19)

tangent of

y = x^{2} + 3, (4, 19)

y = 8 x - 13

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

(4, 19) : m = 8

Find the line with slope m=

8

and passing through

(4, 19) : y = 8 x - 13

y = 8 x - 13

\int \frac{2 x ^{2} + x - 1}{x ( x ^{2} + 1 )} d x

\frac{d}{d y} (\frac{2 x ^{2} - x y + 2 y ^{2}}{x - y})

x^{2} + 2 y^{2} - 2 x - 8 y + 6 = 0 (0.3)

\frac{1}{2 s ^{2} + 1}

\int_{0}^{2} \frac{1}{x ^{2} - 7 x + 6} d x

What is the tangent of y=x^2+3,(4,19) ?
The tangent of y=x^2+3,(4,19) is y=8x-13