Question

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

Short answer

The equation of the line that is tangent to the graph of \(y = e^x\) and goes through the origin is \(y = e(x - 1) + e\).

Step-by-step solution

Calculate the derivative of \(y = e^x\)

The derivative of \(e^x\) with respect to \(x\) is \(e^x\). Hence, the slope of the tangent line at any given point on the graph of \(y = e^x\) is \(e^x\).

Finding the point of tangency

Since we want the tangent line to go through the origin, the slope of the tangent line has to be equal to the y-coordinate of the point of tangency. If we denote this point as (\(x_0, y_0\)), our conditions become \(y_0 = e^{x_0}\) and \(y_0 = e^{x_0}\). Since both conditions have to be fulfilled simultaneously, \(x_0\) must be equal to 1. Thus the point of tangency is (1, \(e\)).

Form the equation of the tangent line

We now can form the equation of the tangent line using the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope of the line and \((x_1, y_1)\) is a point on the line. In this case, \(m = e\), \(x_1 = 1\), and \(y_1 = e\), so the equation becomes \(y - e = e(x - 1)\). This simplifies to \(y = e(x - 1) + e\).