What is the tangent of y=x^2+3x+11 ?

The tangent of y=x^2+3x+11 is y=(2a_{0}+3)x-a_{0}^2+11

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tangent of y=x^2+3x+11

tangent of

y = x^{2} + 3 x + 11

y = (2 a_{0} + 3) x - a_{0}^{2} + 11

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} + 3 a_{0} + 11)

Find the slope of

y = x^{2} + 3 x + 11 : \frac{d y}{d x} = 2 x + 3

Compute the slope of

y = x^{2} + 3 x + 11

x = a_{0} : m = 2 a_{0} + 3

Find the line with slope m=

2 a_{0} + 3

and passing through

(a_{0}, a_{0}^{2} + 3 a_{0} + 11) : y = (2 a_{0} + 3) x - a_{0}^{2} + 11

y = (2 a_{0} + 3) x - a_{0}^{2} + 11

\frac{3}{1 + 3 x}

\int x 5 + 2 x d x

\frac{d}{d x} (2 sin (x) - cos (2 x))

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 e^{3 t}

\frac{d y}{d x} = \frac{sin ( x )}{cos ( y )}

What is the tangent of y=x^2+3x+11 ?
The tangent of y=x^2+3x+11 is y=(2a_{0}+3)x-a_{0}^2+11