What is the tangent of f(x)=25-x^2,(5,0) ?

The tangent of f(x)=25-x^2,(5,0) is y=-10x+50

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tangent of f(x)=25-x^2,(5,0)

tangent of

f (x) = 25 - x^{2}, (5, 0)

y = - 10 x + 50

Solution steps

Find the slope of

f (x) = 25 - x^{2} : \frac{df ( x )}{d x} = - 2 x

Compute the slope of

f (x) = 25 - x^{2}

(5, 0) : m = - 10

Find the line with slope m=

- 10

and passing through

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

y = - 10 x + 50

\frac{x + 7}{3 x ^{2} e ^{x}}

x \to 0 lim (ln (\frac{2}{x}))

y^{^{'}} = (x + y + 7)^{2}

\int \frac{( x ^{2} )}{x ^{3} + 4} d x

\frac{d}{d x} (\frac{8 x ^{2} + 4 x + 6}{x})

What is the tangent of f(x)=25-x^2,(5,0) ?
The tangent of f(x)=25-x^2,(5,0) is y=-10x+50