What is the tangent of f(x)=x^2+3,(-2,7) ?

The tangent of f(x)=x^2+3,(-2,7) is y=-4x-1

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tangent of f(x)=x^2+3,(-2,7)

tangent of

f (x) = x^{2} + 3, (- 2, 7)

y = - 4 x - 1

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

(- 2, 7) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 7) : y = - 4 x - 1

y = - 4 x - 1

x \to 3 lim (\frac{( x ^{2} - 24 x + 25 ) ( x + 5 )}{x - 5})

x \to 7 + lim (\frac{x - 10}{x - 7})

\int \frac{10}{1 - cos ( 5 x )} d x

\int \frac{( x + 1 )}{x ^{2} + 1} d x

\frac{d}{d x} (\frac{2 x}{- 8 x - 1})

What is the tangent of f(x)=x^2+3,(-2,7) ?
The tangent of f(x)=x^2+3,(-2,7) is y=-4x-1