We are asked to find the x-coordinates of the points on the curve $y = x^3 + 7x^2 - 2x + 15$ where the tangent line is parallel to the line $y = -13x + 27$.

Step 1: Identify the slope of the given line

The slope of the line $y = -13x + 27$ is the coefficient of $x$, which is $-13$.

Step 2: Find the derivative of the curve

The derivative of the curve gives the slope of the tangent line at any point on the curve. So, we compute the derivative $\frac{dy}{dx}$ of the function $y = x^3 + 7x^2 - 2x + 15$.

$\frac{dy}{dx} = 3x^2 + 14x - 2$

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

Since we want the tangent line to be parallel to the line $y = -13x + 27$, we set the derivative equal to $-13$:

$3x^2 + 14x - 2 = -13$

Step 4: Solve for $x$

Now, we solve the equation $3x^2 + 14x - 2 = -13$. Simplifying:

$3x^2 + 14x - 2 + 13 = 0$ $3x^2 + 14x + 11 = 0$

Now, solve the quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$, $b = 14$, and $c = 11$. Substituting these values into the quadratic formula:

$x = \frac{-14 \pm \sqrt{14^2 - 4(3)(11)}}{2(3)}$ $x = \frac{-14 \pm \sqrt{196 - 132}}{6}$ $x = \frac{-14 \pm \sqrt{64}}{6}$ $x = \frac{-14 \pm 8}{6}$

So, the two solutions are:

$x = \frac{-14 + 8}{6} = \frac{-6}{6} = -1$ $x = \frac{-14 - 8}{6} = \frac{-22}{6} = -\frac{11}{3}$

Conclusion

The x-coordinates of the points where the tangent line is parallel to the line $y = -13x + 27$ are $x = -1$ and $x = -\frac{11}{3}$.

Would you like more details on this process or a graphical explanation?

5 Related Questions:

1. How can we find the corresponding y-coordinates for these x-values on the curve?
2. What is the meaning of a tangent line being parallel to another line geometrically?
3. How would the solution change if the slope of the given line was different?
4. Can you explain how to solve quadratic equations by factoring instead of using the quadratic formula?
5. What is the significance of the second derivative in analyzing curves?

Tip:

Always check if a quadratic equation can be factored before using the quadratic formula—it often simplifies calculations!