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

x \to 0 + lim (\frac{7}{x} - \frac{7}{∣ x ∣})

\frac{d}{d x} (\frac{2}{1 + x ^{3}})

\frac{s ^{2}}{( s - 3 ) ^{2}}

2 + 5 x^{2} - 2 x^{3}

f (x) = x^{2} - 7 x + 7, at x = 6

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