What is the tangent of y=x^2+5x+9,\at x=4 ?

The tangent of y=x^2+5x+9,\at x=4 is y=13x-7

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tangent of y=x^2+5x+9,\at x=4

tangent of

y = x^{2} + 5 x + 9, at x = 4

y = 13 x - 7

Solution steps

Find the tangent point:

(4, 45)

Find the slope of

y = x^{2} + 5 x + 9 : \frac{d y}{d x} = 2 x + 5

Compute the slope of

y = x^{2} + 5 x + 9

x = 4 : m = 13

Find the line with slope m=

13

and passing through

(4, 45) : y = 13 x - 7

y = 13 x - 7

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

y = 1.2 (x + 0.1)^{0.5} + 4, 1, - 0.1

n = 0 \sum \infty 2^{n} 2^{n}

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

x^{2} - 3 x + 6, at x = 1

What is the tangent of y=x^2+5x+9,\at x=4 ?
The tangent of y=x^2+5x+9,\at x=4 is y=13x-7