Where does the tangent to the curve y = e xat the point (0, 1) meet x-axis?
(-1, 0)
Finding Where the Tangent to y = e^x Meets the x-axis
This question asks us to find the point where the tangent line to the curve \(y = e^x\) at the specific point \((0, 1)\) intersects the x-axis. To do this, we first need to find the equation of the tangent line.
Step 1: Find the Derivative of the Curve
The slope of the tangent line to a curve \(y = f(x)\) at any point is given by its derivative, \(\frac{dy}{dx}\). The curve is given by \(y = e^x\).
The derivative of \(y = e^x\) with respect to \(x\) is:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x \]
Step 2: Calculate the Slope of the Tangent at the Given Point
The given point is \((0, 1)\). To find the slope of the tangent at this point, we substitute the x-coordinate of the point (\(x=0\)) into the derivative we just found.
\[ \text{Slope } (m) = \left(\frac{dy}{dx}\right)_{(0, 1)} = e^0 \]
Recall that any non-zero number raised to the power of 0 is 1. So,
\[ m = 1 \]
The slope of the tangent line at \((0, 1)\) is 1.
Step 3: Find the Equation of the Tangent Line
We have the slope \(m=1\) and a point the line passes through \((x_1, y_1) = (0, 1)\). We can use the point-slope form of the equation of a straight line, which is \(y - y_1 = m(x - x_1)\).
Substituting the values:
\[ y - 1 = 1(x - 0) \]
\[ y - 1 = x \]
Rearranging the equation to the slope-intercept form \(y = mx + c\):
\[ y = x + 1 \]
This is the equation of the tangent line to the curve \(y = e^x\) at the point \((0, 1)\).
Step 4: Find Where the Tangent Line Meets the x-axis
A line meets the x-axis at the point where the y-coordinate is 0. To find this point for our tangent line \(y = x + 1\), we set \(y = 0\):
\[ 0 = x + 1 \]
Now, solve for \(x\):
\[ x = -1 \]
So, the tangent line meets the x-axis at \(x = -1\). The point of intersection is \((-1, 0)\).
Summary of the Solution Process
- Find the derivative \(dy/dx\) of the curve \(y = e^x\).
- Evaluate the derivative at the given point \((0, 1)\) to get the slope \(m\).
- Use the point-slope formula \(y - y_1 = m(x - x_1)\) to get the equation of the tangent line.
- Set \(y=0\) in the tangent line equation and solve for \(x\) to find the x-intercept.
The point where the tangent to the curve \(y = e^x\) at \((0, 1)\) meets the x-axis is \((-1, 0)\).
Revision Table: Key Concepts for Tangent Lines
| Concept | Description | Formula/Method |
|---|---|---|
| Derivative (\(\frac{dy}{dx}\)) | Represents the instantaneous rate of change of \(y\) with respect to \(x\). Gives the slope of the tangent line at any point on the curve. | \(\frac{dy}{dx} = f'(x)\) |
| Slope of Tangent | The specific value of the derivative at a given point \((x_1, y_1)\). | \(m = f'(x_1)\) |
| Point-Slope Form | Equation of a line given its slope \(m\) and a point \((x_1, y_1)\) it passes through. | \(y - y_1 = m(x - x_1)\) |
| x-intercept | The point where a line crosses the x-axis. At this point, the y-coordinate is always 0. | Set \(y = 0\) in the line equation and solve for \(x\). The point is \((x, 0)\). |
Additional Information: The Exponential Function y = e^x
The function \(y = e^x\) is the natural exponential function. Its base is the mathematical constant \(e\), which is approximately 2.71828. It has some unique properties:
- The derivative of \(e^x\) is \(e^x\) itself (\(\frac{d}{dx}(e^x) = e^x\)). This property makes it fundamental in calculus.
- The graph of \(y = e^x\) always passes through the point \((0, 1)\) because \(e^0 = 1\).
- The graph is always positive (\(e^x > 0\) for all real \(x\)).
- The graph increases rapidly as \(x\) increases and approaches 0 as \(x\) approaches negative infinity.
- The y-axis is the tangent line at \((0,1)\) only for vertical tangent cases, but here we are finding the tangent to the *curve* at \((0,1)\) and where *that* tangent meets the x-axis.
Understanding the basic properties of functions like \(y = e^x\) is helpful when dealing with calculus problems involving them.
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