The Straight Line Tangent to yx^3 at xa Crosses the Y-Axis at (0,2): Analyzing the Tangent Line
To find the value of a such that the tangent line to the curve y x^3 at xa crosses the y-axis at the point (0,2), we can follow a systematic approach that involves finding the slope of the tangent line, using the point-slope form, and then solving for the value of a.
1. Finding the Slope of the Tangent Line
The first step is to find the derivative of y x^3, which gives us the slope of the tangent line at any point on the curve.
Step 1: Find the derivative of y x^3:
Mathematically, (frac{dy}{dx} 3x^2)
Step 2: Evaluate the derivative at xa to find the slope of the tangent line:
(m 3a^2)
2. Finding the Point of Tangency
Next, we need to find the point on the curve where x a.
Step 3: Find the y-coordinate of the point of tangency by substituting a into y x^3: (y a^3)
Thus, the point of tangency is (a, a^3).
3. Using the Point-Slope Form of the Equation of the Tangent Line
With the point of tangency and the slope, we can use the point-slope form of the equation of a line.
The point-slope form is given by (y - y_1 m(x - x_1)), where (x_1, y_1) is the point of tangency and m is the slope.
Substituting the values, we get:
(y - a^3 3a^2(x - a))
Simplifying this, we get:
(y - a^3 3a^2x - 3a^3)
(y 3a^2x - 2a^3)
4. Finding the Y-Intercept
The tangent line crosses the y-axis at (0, 2). To find the y-intercept, we set x 0 in the equation of the tangent line.
Substituting x 0 into the equation:
(y 3a^2(0) - 2a^3)
(y -2a^3)
We know that the y-intercept is 2, so we set (-2a^3 2).
Solving for a:
(a^3 -1)
(a -1)
Conclusion
Thus, the value of a is (a -1).
Example Verification
To verify, we can check if the tangent line at x -1 crosses the y-axis at (0, 2).
The slope at x -1 is (3(-1)^2 3).
The point of tangency is (-1, (-1)^3) or (-1, -1).
The equation of the tangent line at this point is:
(y 1 3(x 1))
(y 3x 2)
Setting x 0, we get y 2, which confirms that the line crosses the y-axis at (0, 2).
Therefore, the value of a is (boxed{-1}).