What is the tangent of y=5+4x^2 ?

The tangent of y=5+4x^2 is y=8a_{0}x+5-4a_{0}^2

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tangent of y=5+4x^2

tangent of

y = 5 + 4 x^{2}

y = 8 a_{0} x + 5 - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 + 4 a_{0}^{2})

Find the slope of

y = 5 + 4 x^{2} : \frac{d y}{d x} = 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8 a_{0}

Find the line with slope m=

8 a_{0}

and passing through

(a_{0}, 5 + 4 a_{0}^{2}) : y = 8 a_{0} x + 5 - 4 a_{0}^{2}

y = 8 a_{0} x + 5 - 4 a_{0}^{2}

\frac{\partial}{\partial x} (12 x^{2} y^{3})

x \to 0 + lim ((e^{x} + x)^{\frac{3}{x}})

\int \frac{1}{x ^{2} + 2 x + 3} d x

\frac{d y}{d x} = \frac{x}{1 + 2 y}

f (x) = \frac{x ^{3}}{3} + 2 x^{2} + 13 x - 5

What is the tangent of y=5+4x^2 ?
The tangent of y=5+4x^2 is y=8a_{0}x+5-4a_{0}^2