What is the tangent of f(x)=x^2-3,(-4,13) ?

The tangent of f(x)=x^2-3,(-4,13) is y=-8x-19

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tangent of f(x)=x^2-3,(-4,13)

tangent of

f (x) = x^{2} - 3, (- 4, 13)

y = - 8 x - 19

Solution steps

Find the slope of

f (x) = x^{2} - 3 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 3

(- 4, 13) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 13) : y = - 8 x - 19

y = - 8 x - 19

\int \frac{x ^{2} - 1}{x ^{3} + x} d x

\int_{0}^{1} x e^{3 x^{2}} d x

tan (t) y^{^{'}} - y = 1, y (\frac{π}{6}) = 3

\frac{x}{x ^{2} - x + 1}

y^{^{''}} - 2 y^{^{'}} + y = e^{t}

What is the tangent of f(x)=x^2-3,(-4,13) ?
The tangent of f(x)=x^2-3,(-4,13) is y=-8x-19