What is the tangent of y=2x^2+1,(3,19) ?

The tangent of y=2x^2+1,(3,19) is y=12x-17

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tangent of y=2x^2+1,(3,19)

tangent of

y = 2 x^{2} + 1, (3, 19)

y = 12 x - 17

Solution steps

Find the slope of

y = 2 x^{2} + 1 : \frac{d y}{d x} = 4 x

Compute the slope of

y = 2 x^{2} + 1

(3, 19) : m = 12

Find the line with slope m=

12

and passing through

(3, 19) : y = 12 x - 17

y = 12 x - 17

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

\frac{x ^{3}}{2 - x ^{2}}

(sin^{4} (x))^{^{'}}

\frac{2}{( s + 1 ) ^{3}}

\frac{d}{d x} (x e^{\frac{y}{x}})

What is the tangent of y=2x^2+1,(3,19) ?
The tangent of y=2x^2+1,(3,19) is y=12x-17