# Thread: [SOLVED] Find equations of the tangent line...

1. ## [SOLVED] Find equations of the tangent line...

Find equations of the tangent line and normal line to the given curve at the specified point.

Y = 2xe^(x) , (0,0)

2. Originally Posted by moonman
Find equations of the tangent line and normal line to the given curve at the specified point.

Y = 2xe^(x) , (0,0)
A tangent line is a straight line that touches a function at only one point. The slope of the tangent line (the gradient) at a point on the function is equal to the derivative of the function at the same point.

$\displaystyle y=2xe^x$

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 2xe^x + 2e^x$. At $\displaystyle (0,0)$, The tangent gradient equals $\displaystyle m = 2(0)e^0 + 2e^0 = 2$

Equation of tangent: $\displaystyle y-0 = 2(x-0) \rightarrow y = 2x$

The equation of Normal line has the same coordinate of intersection $\displaystyle (0,0)$ and the same y-intercept but the gradient is different. The gradient of the normal line is $\displaystyle n = -\frac{1}{m} = -\frac{1}{2}$ where $\displaystyle m$ is the gradient of the tangent line and $\displaystyle n$ is the gradient of the normal line.

This Equation of normal line: $\displaystyle y-0 = -\frac12 (x-0) \rightarrow y = -\frac12 x$

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### slope of normal of the carve formula of math

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