We are tasked with finding the points on the curve $y = x^2 + x - 2$ where the tangent line is parallel to the line $y = 4x - 1$.

Step 1: Find the slope of the given line

The line $y = 4x - 1$ is in slope-intercept form $y = mx + b$, where the slope $m$ is 4.

Step 2: Find the derivative of the curve

To find the points where the tangent line to the curve is parallel to $y = 4x - 1$, we need to compute the derivative of the curve $y = x^2 + x - 2$, which will give us the slope of the tangent line at any point $x$.

$\frac{dy}{dx} = 2x + 1$

Step 3: Set the derivative equal to the slope of the line

For the tangent to be parallel to the line $y = 4x - 1$, the slope of the tangent must be equal to 4. So we set the derivative equal to 4:

$2x + 1 = 4$

Step 4: Solve for $x$

$2x = 4 - 1$ $2x = 3$ $x = \frac{3}{2}$

Step 5: Find the corresponding $y$-coordinate

Now that we have $x = \frac{3}{2}$, we substitute this back into the original curve equation $y = x^2 + x - 2$ to find the corresponding $y$-coordinate.

$y = \left( \frac{3}{2} \right)^2 + \frac{3}{2} - 2$ $y = \frac{9}{4} + \frac{3}{2} - 2$ $y = \frac{9}{4} + \frac{6}{4} - \frac{8}{4}$ $y = \frac{7}{4}$

Final answer:

The point on the curve where the tangent line is parallel to the line $y = 4x - 1$ is $\left( \frac{3}{2}, \frac{7}{4} \right)$.

Would you like more details on any specific step?

Here are 5 related questions:

1. How do you find the slope of a tangent line at any point on a curve?
2. Can you explain the significance of the derivative in finding tangent lines?
3. What happens when a tangent line has the same slope as a given line?
4. How can we apply this method to other curves, such as $y = x^3 - x$?
5. How would the problem change if the given line were $y = -4x + 1$?

Tip: Always check the derivative to ensure the slope matches the given condition when solving problems involving tangents.