What is the tangent of f(x)=x^2+8,(-2,12) ?

The tangent of f(x)=x^2+8,(-2,12) is y=-4x+4

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tangent of f(x)=x^2+8,(-2,12)

tangent of

f (x) = x^{2} + 8, (- 2, 12)

y = - 4 x + 4

Solution steps

Find the slope of

f (x) = x^{2} + 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 8

(- 2, 12) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 12) : y = - 4 x + 4

y = - 4 x + 4

\frac{d}{d y} (- \frac{y}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

\int (3 x^{2} - x) d x

\frac{d}{d x} (e^{3 - 2 x})

\int (x^{2} + 1)^{9} d x

What is the tangent of f(x)=x^2+8,(-2,12) ?
The tangent of f(x)=x^2+8,(-2,12) is y=-4x+4