What is the tangent of y=(1+2x)^2,(3,49) ?

The tangent of y=(1+2x)^2,(3,49) is y=28x-35

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tangent of y=(1+2x)^2,(3,49)

tangent of

y = (1 + 2 x)^{2}, (3, 49)

y = 28 x - 35

Solution steps

Find the slope of

y = (1 + 2 x)^{2} : \frac{d y}{d x} = 4 (1 + 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 = 28

Find the line with slope m=

28

and passing through

(3, 49) : y = 28 x - 35

y = 28 x - 35

What is the tangent of y=(1+2x)^2,(3,49) ?
The tangent of y=(1+2x)^2,(3,49) is y=28x-35