What is the tangent of f(x)=x^2-7,(-4,9) ?

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

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tangent of f(x)=x^2-7,(-4,9)

tangent of

f (x) = x^{2} - 7, (- 4, 9)

y = - 8 x - 23

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, 9) : m = - 8

Find the line with slope m=

- 8

and passing through

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

y = - 8 x - 23

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

\int y cos (x y) d x

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

\frac{d}{d x} (3 sec^{2} (x))

\frac{d y}{d x} = cos (5 x)

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