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

The tangent of f(x)=5x^2,\at x=-2 is y=-20x-20

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

tangent of

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

y = - 20 x - 20

Solution steps

Find the tangent point:

(- 2, 20)

Find the slope of

f (x) = 5 x^{2} : \frac{df ( x )}{d x} = 10 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 = - 20

Find the line with slope m=

- 20

and passing through

(- 2, 20) : y = - 20 x - 20

y = - 20 x - 20

y = t (ln (4 t))^{2}

h \to 0 lim (\frac{2 ( x + h ) - 2 x}{h})

\int \frac{1}{- y} d y

\int \frac{e ^{y} x ^{2}}{y}

\int (x^{2} + 1 + x)^{4} d x

What is the tangent of f(x)=5x^2,\at x=-2 ?
The tangent of f(x)=5x^2,\at x=-2 is y=-20x-20