What is the tangent of y=x^2+3,(-5,28) ?

The tangent of y=x^2+3,(-5,28) is y=-10x-22

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tangent of y=x^2+3,(-5,28)

tangent of

y = x^{2} + 3, (- 5, 28)

y = - 10 x - 22

Solution steps

Find the slope of

y = x^{2} + 3 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 3

(- 5, 28) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 28) : y = - 10 x - 22

y = - 10 x - 22

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

\int x^{2} a^{2} + a^{2} x^{2} d x

\frac{d}{d x} (sec (x) + tan (x))

\frac{13}{x}

f (x) = 2 x^{4} - 6 x^{2}, at x = 1

What is the tangent of y=x^2+3,(-5,28) ?
The tangent of y=x^2+3,(-5,28) is y=-10x-22