What is the tangent of f(x)=-5x^2,\at a=4 ?

The tangent of f(x)=-5x^2,\at a=4 is y=-10ax+5a^2

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tangent of f(x)=-5x^2,\at a=4

tangent of

f (x) = - 5 x^{2}, at a = 4

y = - 10 a x + 5 a^{2}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, - 5 a^{2})

Find the slope of

f (x) = - 5 x^{2} : \frac{df ( x )}{d x} = - 10 x

Compute the slope of

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

x = a : m = - 10 a

Find the line with slope m=

- 10 a

and passing through

(a, - 5 a^{2}) : y = - 10 a x + 5 a^{2}

y = - 10 a x + 5 a^{2}

y^{^{''}} - (\frac{2}{x}) y^{^{'}} - (\frac{7}{x ^{2}}) y = 0

x \to 1 lim (51 - 2 x)

\frac{d}{d x} (ln (sec (x^{2})))

\int x^{\frac{3}{4}} y^{3} d y

\int e^{x + y^{2}} d x

What is the tangent of f(x)=-5x^2,\at a=4 ?
The tangent of f(x)=-5x^2,\at a=4 is y=-10ax+5a^2