What is the tangent of f(x)=x^2+7,(-4,23) ?

The tangent of f(x)=x^2+7,(-4,23) is y=-8x-9

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tangent of f(x)=x^2+7,(-4,23)

tangent of

f (x) = x^{2} + 7, (- 4, 23)

y = - 8 x - 9

Solution steps

Find the slope of

f (x) = x^{2} + 7 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 7

(- 4, 23) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 23) : y = - 8 x - 9

y = - 8 x - 9

x \to 1 + lim (x^{\frac{4}{1 - x}})

\int \frac{- 2 x - 6}{x ^{3} - 2 x ^{2} - 3 x} d x

x \to 1 + lim (\frac{4}{x - 1})

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

x \to 0 lim (x^{2 x})

What is the tangent of f(x)=x^2+7,(-4,23) ?
The tangent of f(x)=x^2+7,(-4,23) is y=-8x-9