What is the tangent of f(x)=5-8x^2 ?

The tangent of f(x)=5-8x^2 is y=-16a_{0}x+5+8a_{0}^2

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tangent of f(x)=5-8x^2

tangent of

f (x) = 5 - 8 x^{2}

y = - 16 a_{0} x + 5 + 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 - 8 a_{0}^{2})

Find the slope of

f (x) = 5 - 8 x^{2} : \frac{df ( x )}{d x} = - 16 x

Compute the slope of

f (x) = 5 - 8 x^{2}

x = a_{0} : m = - 16 a_{0}

Find the line with slope m=

- 16 a_{0}

and passing through

(a_{0}, 5 - 8 a_{0}^{2}) : y = - 16 a_{0} x + 5 + 8 a_{0}^{2}

y = - 16 a_{0} x + 5 + 8 a_{0}^{2}

\frac{d y}{d x} = 7 x y

\frac{d}{d x} (e^{3 x + 2 y})

f (x) = (7 x - 8)^{2}

\frac{d}{d y} (y - x)

\int (2 e^{x} + 4^{x}) d x

What is the tangent of f(x)=5-8x^2 ?
The tangent of f(x)=5-8x^2 is y=-16a_{0}x+5+8a_{0}^2