What is the tangent of y=x^2+4x+7,\at x=1 ?

The tangent of y=x^2+4x+7,\at x=1 is y=6x+6

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tangent of y=x^2+4x+7,\at x=1

tangent of

y = x^{2} + 4 x + 7, at x = 1

y = 6 x + 6

Solution steps

Find the tangent point:

(1, 12)

Find the slope of

y = x^{2} + 4 x + 7 : \frac{d y}{d x} = 2 x + 4

Compute the slope of

y = x^{2} + 4 x + 7

x = 1 : m = 6

Find the line with slope m=

6

and passing through

(1, 12) : y = 6 x + 6

y = 6 x + 6

\frac{7}{2 7 x + 1}, at x = 9

ln (y^{2})

\frac{d}{d x} (arctan (1 - x))

\int_{2}^{3} 2 x d x

\frac{7}{4} x^{4} - \frac{6}{5} x^{3}

What is the tangent of y=x^2+4x+7,\at x=1 ?
The tangent of y=x^2+4x+7,\at x=1 is y=6x+6