What is the tangent of f(x)=x^2+1,(-4,17) ?

The tangent of f(x)=x^2+1,(-4,17) is y=-8x-15

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tangent of f(x)=x^2+1,(-4,17)

tangent of

f (x) = x^{2} + 1, (- 4, 17)

y = - 8 x - 15

Solution steps

Find the slope of

f (x) = x^{2} + 1 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} + 1

(- 4, 17) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 17) : y = - 8 x - 15

y = - 8 x - 15

2 x y + x^{2} \frac{d y}{d x} = y^{2}

\frac{z ^{\frac{8}{3}}}{8} - \frac{e ^{- 2 z}}{2}

\frac{d}{d x} (\frac{3 x + 4}{2 x - 5})

y^{^{''}} - 3 y^{^{'}} = 0

x \to \infty lim (7 x + 4)

What is the tangent of f(x)=x^2+1,(-4,17) ?
The tangent of f(x)=x^2+1,(-4,17) is y=-8x-15