What is the tangent of y=x^2+x,(4,20) ?

The tangent of y=x^2+x,(4,20) is y=9x-16

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tangent of y=x^2+x,(4,20)

tangent of

y = x^{2} + x, (4, 20)

y = 9 x - 16

Solution steps

Find the slope of

y = x^{2} + x : \frac{d y}{d x} = 2 x + 1

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 = 9

Find the line with slope m=

9

and passing through

(4, 20) : y = 9 x - 16

y = 9 x - 16

(2 y^{2} + 4 x^{2}) d x = x y d y

g (u) = 10 u + 7 u

x (\frac{d y}{d x}) + y = e^{x}

x \to 3 lim (- \frac{x - 4}{x - 3})

t^{2} - \frac{1}{t ^{3}}

What is the tangent of y=x^2+x,(4,20) ?
The tangent of y=x^2+x,(4,20) is y=9x-16