What is the tangent of y=x^2+5,(4,21) ?

The tangent of y=x^2+5,(4,21) is y=8x-11

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tangent of y=x^2+5,(4,21)

tangent of

y = x^{2} + 5, (4, 21)

y = 8 x - 11

Solution steps

Find the slope of

y = x^{2} + 5 : \frac{d y}{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 = 8

Find the line with slope m=

8

and passing through

(4, 21) : y = 8 x - 11

y = 8 x - 11

u \to 0 lim (ln (\frac{u + 1}{u}))

y = - x^{3} + 2 x^{2} - 4, at x = - 1

(- 4.3) (5.3)

n = 1 \sum \infty \frac{5}{3 ^{n} + 1}

n = 0 \sum \infty (2 n - 3) (4 x)^{n}

What is the tangent of y=x^2+5,(4,21) ?
The tangent of y=x^2+5,(4,21) is y=8x-11