What is the tangent of f(x)=x^2+6,(4,22) ?

The tangent of f(x)=x^2+6,(4,22) is y=8x-10

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tangent of f(x)=x^2+6,(4,22)

tangent of

f (x) = x^{2} + 6, (4, 22)

y = 8 x - 10

Solution steps

Find the slope of

f (x) = x^{2} + 6 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 6

(4, 22) : m = 8

Find the line with slope m=

8

and passing through

(4, 22) : y = 8 x - 10

y = 8 x - 10

(1, 6), (0.5, 8)

x \to 0 lim ((1 - 2 x)^{\frac{1}{2}})

\frac{150}{s ( s ^{2} + 8 s + 25 )}

\frac{\partial}{\partial x} (\frac{1}{f ( x )})

\frac{d}{d x} (\frac{4 x}{7 - cot ( x )})

What is the tangent of f(x)=x^2+6,(4,22) ?
The tangent of f(x)=x^2+6,(4,22) is y=8x-10