What is the tangent of f(x)=x^2-4,(3,5) ?

The tangent of f(x)=x^2-4,(3,5) is y=6x-13

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tangent of f(x)=x^2-4,(3,5)

tangent of

f (x) = x^{2} - 4, (3, 5)

y = 6 x - 13

Solution steps

Find the slope of

f (x) = x^{2} - 4 : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6

Find the line with slope m=

6

and passing through

(3, 5) : y = 6 x - 13

y = 6 x - 13

\frac{d}{d x} (2 x^{6} - 5 x^{- \frac{1}{2}})

sin^{2} (π t)

n = 1 \sum \infty n

\int_{- \infty}^{0} \frac{1}{4 - 7 x} d x

t \to - 2 lim (\frac{t + 2}{t ^{3} + 8})

What is the tangent of f(x)=x^2-4,(3,5) ?
The tangent of f(x)=x^2-4,(3,5) is y=6x-13