What is the tangent of y=x^2+2x-1,(1,2) ?

The tangent of y=x^2+2x-1,(1,2) is y=4x-2

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tangent of y=x^2+2x-1,(1,2)

tangent of

y = x^{2} + 2 x - 1, (1, 2)

y = 4 x - 2

Solution steps

Find the slope of

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

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

(1, 2) : y = 4 x - 2

y = 4 x - 2

x, y = 2, x = 0

\frac{\partial}{\partial x} (sin (x) + cos (y))

f (x) = \frac{x ^{3}}{x - 1}

n = 0 \sum \infty (\frac{2}{7})^{\frac{n}{2}}

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 0

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