What is the tangent of f(x)=-0.5x^2,\at x=10 ?

The tangent of f(x)=-0.5x^2,\at x=10 is y=-10x+50

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tangent of f(x)=-0.5x^2,\at x=10

tangent of

f (x) = - 0.5 x^{2}, at x = 10

y = - 10 x + 50

Solution steps

Find the tangent point:

(10, - 50)

Find the slope of

f (x) = - 0.5 x^{2} : \frac{df ( x )}{d x} = - 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 = - 10

Find the line with slope m=

- 10

and passing through

(10, - 50) : y = - 10 x + 50

y = - 10 x + 50

\int_{0}^{1} 4 (1 + x)^{5} d x

y^{^{''}} + y = 3 csc^{2} (t)

\frac{\partial}{\partial x} (\frac{6 y}{x ^{2} + y ^{2} + 5})

\frac{d}{d x} ((4 x - 5)^{3})

\int \frac{1}{e ^{- 3 x}} d x

What is the tangent of f(x)=-0.5x^2,\at x=10 ?
The tangent of f(x)=-0.5x^2,\at x=10 is y=-10x+50