Final answer:
The tangent to the curve y=e^{2x} at the point (0,1) meets the x-axis at the point (-1/2, 0). To determine this, the derivative of the curve was used to find the slope of the tangent, which was then used to find the tangent line equation. Setting the y-coordinate to zero gave the x-intercept of the tangent line.
Step-by-step explanation:
The student is asking about the point where the tangent to the curve y=e^{2x} at the point (0,1) intersects the x-axis. To find the point of intersection, we first need to determine the slope of the tangent to the curve at that point. The slope of a curve at a specific point is given by its derivative at that point.
We can calculate the derivative of y=e^{2x} as follows:
y' = d/dx (e^{2x}) = 2e^{2x}.
At the point (0,1), the slope of the tangent is therefore y'(0) = 2e^{2(0)} = 2.
The equation of the tangent line can be written in the slope-intercept form as:
y = mx + c,
where m is the slope and c is the y-intercept. Since the line is tangent to the curve at (0,1), the point (0,1) lies on the line, which gives us the equation:
y = 2x + 1.
To find the x-coordinate of the point where this line meets the x-axis, we need to set y to 0 (since any point on the x-axis has a y-coordinate of 0) and solve for x:
0 = 2x + 1,
x = -1/2.
Therefore, the tangent meets the x-axis at the point (-1/2, 0).