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

Compute the slope of

f (x) = - 0.5 x^{2}

x = 10 : 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

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

\frac{\partial}{\partial x} (x^{6} y - 4 x^{3} y^{2})

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

x \to 0 lim (- \frac{1}{x - 1})

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