What is the tangent of f(x)=x^2+5,(2,9) ?

The tangent of f(x)=x^2+5,(2,9) is y=4x+1

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tangent of f(x)=x^2+5,(2,9)

tangent of

f (x) = x^{2} + 5, (2, 9)

y = 4 x + 1

Solution steps

Find the slope of

f (x) = x^{2} + 5 : \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 = 4

Find the line with slope m=

4

and passing through

(2, 9) : y = 4 x + 1

y = 4 x + 1

10 1 + x

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

\frac{\partial}{\partial t} (e^{- 7 t} cos (π x))

x \to \frac{π}{4} lim (\frac{sin ( x - \frac{π}{4} )}{4 x - π})

x^{^{'}} + x = 0

What is the tangent of f(x)=x^2+5,(2,9) ?
The tangent of f(x)=x^2+5,(2,9) is y=4x+1