Key Concepts and Formulas

• Derivative as Slope of Tangent: The derivative of a function $f(x)$ at a point $x_0$, denoted as $f'(x_0)$, gives the slope of the tangent line to the curve $y = f(x)$ at the point $(x_0, f(x_0))$.
• Equation of a Tangent Line (Point-Slope Form): The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is given by $y - y_1 = m(x - x_1)$.
• Point of Tangency Condition: A point is a point of tangency if it lies on both the curve and the tangent line, and the slope of the curve at that point equals the slope of the tangent line.

Step-by-Step Solution

Step 1: Find the equation of the tangent line to the second curve at (2, -1)

• We are given the curve $y = 5x^2 + 2x - 25$ and the point of tangency $(2, -1)$. We need to find the equation of the tangent line to this curve at this point.
• Step 1.1: Calculate the derivative of the second curve. To find the slope of the tangent line, we differentiate $y = 5x^2 + 2x - 25$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(5x^2 + 2x - 25) = 10x + 2$ This derivative gives us the slope of the tangent at any point on the curve.
• Step 1.2: Evaluate the derivative at x = 2 to find the slope at (2, -1). Substituting $x = 2$ into the derivative, we get the slope $m$ of the tangent line at the point $(2, -1)$: $m = 10(2) + 2 = 20 + 2 = 22$ So, the slope of the tangent line at $(2, -1)$ is $22$.
• Step 1.3: Write the equation of the tangent line using the point-slope form. Using the point-slope form, $y - y_1 = m(x - x_1)$, with $(x_1, y_1) = (2, -1)$ and $m = 22$, we get: $y - (-1) = 22(x - 2)$ $y + 1 = 22x - 44$ $y = 22x - 45$ This is the equation of the tangent line to the second curve at the point $(2, -1)$.

Step 2: Relate the tangent line to the first curve

• We are given the curve $y = x^3 - x^2 + x$ and the point of tangency $(a, b)$. We know that the tangent line we found in Step 1 is also tangent to this curve at $(a, b)$.
• Step 2.1: Calculate the derivative of the first curve. To find the slope of the tangent line to the first curve at $(a, b)$, we differentiate $y = x^3 - x^2 + x$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^3 - x^2 + x) = 3x^2 - 2x + 1$ This derivative gives us the slope of the tangent at any point on the curve.
• Step 2.2: Evaluate the derivative at x = a and equate it to the slope of the tangent line. The slope of the tangent line to the first curve at $(a, b)$ is $3a^2 - 2a + 1$. Since this tangent line is the same as the one we found in Step 1, its slope must also be $22$. Therefore: $3a^2 - 2a + 1 = 22$

Step 3: Solve for the point of tangency (a, b)

• Step 3.1: Solve the quadratic equation for a. Rearrange the equation from Step 2.2: $3a^2 - 2a - 21 = 0$ We can solve this quadratic equation by factoring: $3a^2 - 9a + 7a - 21 = 0$ $3a(a - 3) + 7(a - 3) = 0$ $(3a + 7)(a - 3) = 0$ This gives two possible values for $a$: $a = 3 \quad \text{or} \quad a = -\frac{7}{3}$

• Step 3.2: Find the corresponding b values for each a. Since $(a, b)$ is a point on the curve $y = x^3 - x^2 + x$, we have $b = a^3 - a^2 + a$.

  • Case 1: $a = 3$ $b = (3)^3 - (3)^2 + (3) = 27 - 9 + 3 = 21$ So, $(a, b) = (3, 21)$. We need to check if this point lies on the tangent line $y = 22x - 45$. $21 = 22(3) - 45 = 66 - 45 = 21$ Since the equation holds, $(3, 21)$ is a valid point of tangency.

  • Case 2: $a = -\frac{7}{3}$ $b = \left(-\frac{7}{3}\right)^3 - \left(-\frac{7}{3}\right)^2 + \left(-\frac{7}{3}\right) = -\frac{343}{27} - \frac{49}{9} - \frac{7}{3} = -\frac{343}{27} - \frac{147}{27} - \frac{63}{27} = -\frac{553}{27}$ So, $(a, b) = \left(-\frac{7}{3}, -\frac{553}{27}\right)$. We need to check if this point lies on the tangent line $y = 22x - 45$. $-\frac{553}{27} = 22\left(-\frac{7}{3}\right) - 45 = -\frac{154}{3} - 45 = -\frac{154}{3} - \frac{135}{3} = -\frac{289}{3} = -\frac{2601}{27}$ Since $-\frac{553}{27} \neq -\frac{2601}{27}$, this point is not on the tangent line and is not a valid point of tangency.

• Therefore, the only valid point of tangency is $(a, b) = (3, 21)$.

Step 4: Calculate |2a + 9b|

• We have $a = 3$ and $b = 21$. We need to find $|2a + 9b|$. $|2a + 9b| = |2(3) + 9(21)| = |6 + 189| = |195| = 195$

There appears to be an error in the problem statement or the provided correct answer. Let's re-examine the calculations.

Upon closer inspection, the error must be in the point (2, -1) lying on the curve $y = 5x^2 + 2x - 25$. If $x = 2$, then $y = 5(2^2) + 2(2) - 25 = 20 + 4 - 25 = -1$. So, the point is indeed on the curve.

The problem is that the provided answer is wrong. Let's check if the tangent to the first curve at (3, 21) is tangent to the second curve. The equation of the tangent is $y = 22x - 45$. Substitute this into the second equation to find intersection points: $22x - 45 = 5x^2 + 2x - 25$ $5x^2 - 20x + 20 = 0$ $x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0$ So, $x = 2$.

Thus, the tangent to the first curve at (3, 21) IS the tangent to the second curve at (2, -1).

|2a + 9b| = |2(3) + 9(21)| = |6 + 189| = |195| = 195

Common Mistakes & Tips

• Careless Differentiation: Ensure you correctly differentiate the given equations. A small error in differentiation will lead to an incorrect slope and subsequent calculations.
• Verification of Tangency: Always verify that the point of tangency lies on both the curve and the tangent line.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when solving quadratic equations and substituting values.

Summary

We found the equation of the tangent line to the second curve at the given point. Then, we found the derivative of the first curve and equated it to the slope of the tangent line. We solved for the x-coordinate of the point of tangency on the first curve and found the corresponding y-coordinate. Finally, we calculated $|2a + 9b|$. The correct value is 195.

Final Answer

The final answer is \boxed{195}.