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)

Compute the slope of

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

(3, 49) : m = 28

Find the line with slope m=

28

and passing through

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

y = 28 x - 35

\frac{s}{( s + 1 ) ^{2} + 4}

\frac{\partial}{\partial y} (x^{2} + y^{2} + z^{2})

\int 5 cos^{3} (2 x) d x

\int \frac{ln ( x ^{3} )}{- 804 x ^{2}} d x

(0.3, - 1), (- 0.1, - 0.8)

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