Question

Find an equation for a line that is normal to the graph of \(y=x e^{x}\)and goes through the origin

Short answer

The equation of the line normal to the graph of \(y = xe^x\) that passes through the origin is \(y = -x\).

Step-by-step solution

Differentiate the Function

Start by differentiating the function \(y = xe^x\) using the product rule for differentiation, which states that the derivative of two functions multiplied together is the first function multiplied by the derivative of the second, plus the second function multiplied by the derivative of the first. In this case, the first function is \(x\) and the second function is \(e^x\). So, taking derivatives: \(y' = e^x + xe^x\).

Find the Slope of the Function at the Origin

Plug the x-coordinate of the origin into the derivative to find the slope of the tangent line at this point. The origin has x-coordinate 0, so our slope is: \(y'(0) = e^0 + 0*e^0 = 1 + 0 = 1\).

Calculate the Slope of Normal

Since a normal line is perpendicular to a tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. The negative reciprocal of 1 is -1; hence the slope of the normal line would be -1.

Determine the Equation of the Line

Use the slope-intercept form of a linear equation (i.e., \(y = mx + c\)), with the slope \(m\) found in the previous step and the y-intercept \(c\) is zero (since the line passes through the origin), to find the equation of the line. Hence the equation of this line is \(y = -x + 0\) or simply, \(y = -x\).