Key Concepts and Formulas

• Equation of a Line (Point-Slope Form): $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.
• Derivative as Slope: The derivative $f'(x)$ of a function $f(x)$ gives the slope of the tangent line to the curve $y = f(x)$ at the point $x$.
• Normal Line: The normal line to a curve at a point is perpendicular to the tangent line at that point. If the tangent line has slope $m$, the normal line has slope $-1/m$.

Step-by-Step Solution

Step 1: Determine the Equation of the Quadratic Curve $y=f(x)$

We are given that the quadratic curve passes through $(-1,0)$ and touches the line $y=x$ at $(1,1)$. This means the line $y=x$ is tangent to the curve at $(1,1)$. Let the quadratic be $f(x) = ax^2 + bx + c$.

• Condition 1: Passes through (-1, 0): Substituting $x = -1$ and $y = 0$ into the equation gives $0 = a(-1)^2 + b(-1) + c \Rightarrow a - b + c = 0 \quad \text{(Equation 1)}$

• Condition 2: Passes through (1, 1): Substituting $x = 1$ and $y = 1$ into the equation gives $1 = a(1)^2 + b(1) + c \Rightarrow a + b + c = 1 \quad \text{(Equation 2)}$

• Condition 3: Tangent to $y = x$ at (1, 1): The derivative of $f(x)$ is $f'(x) = 2ax + b$. The slope of the tangent line $y = x$ is 1. Therefore, $f'(1) = 1$. $f'(1) = 2a(1) + b = 2a + b = 1 \quad \text{(Equation 3)}$

Now we have a system of three equations with three unknowns: $a - b + c = 0$ $a + b + c = 1$ $2a + b = 1$

Subtracting Equation 1 from Equation 2 gives: $(a + b + c) - (a - b + c) = 1 - 0 \Rightarrow 2b = 1 \Rightarrow b = \frac{1}{2}$

Substituting $b = \frac{1}{2}$ into Equation 3 gives: $2a + \frac{1}{2} = 1 \Rightarrow 2a = \frac{1}{2} \Rightarrow a = \frac{1}{4}$

Substituting $a = \frac{1}{4}$ and $b = \frac{1}{2}$ into Equation 2 gives: $\frac{1}{4} + \frac{1}{2} + c = 1 \Rightarrow \frac{3}{4} + c = 1 \Rightarrow c = \frac{1}{4}$

Therefore, the quadratic curve is: $f(x) = \frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{4} = \frac{1}{4}(x^2 + 2x + 1) = \frac{1}{4}(x+1)^2$

Step 2: Find the Equation of the Normal Line at $(\alpha, \alpha+1)$

We are given that the point is $(\alpha, \alpha+1)$ and lies on the curve $y=f(x)$. But $y = f(x) = \frac{1}{4}(x+1)^2$. So, $\alpha + 1 = \frac{1}{4}(\alpha+1)^2$. Since we are in the first quadrant $\alpha+1 > 0$, then we can divide by $(\alpha+1)$.

$1 = \frac{1}{4}(\alpha+1) \Rightarrow 4 = \alpha+1 \Rightarrow \alpha = 3$

So, the point is $(3, 4)$.

Now, we need to find the slope of the tangent at $x=3$. The derivative is $f'(x) = \frac{1}{2}(x+1)$. $f'(3) = \frac{1}{2}(3+1) = \frac{1}{2}(4) = 2$

The slope of the normal line at $(3, 4)$ is the negative reciprocal of the tangent's slope: $m_{normal} = -\frac{1}{2}$

The equation of the normal line at $(3, 4)$ is: $y - 4 = -\frac{1}{2}(x - 3)$

Step 3: Find the x-intercept of the Normal Line

To find the x-intercept, set $y = 0$ in the equation of the normal line: $0 - 4 = -\frac{1}{2}(x - 3)$ $-4 = -\frac{1}{2}(x - 3)$ $8 = x - 3$ $x = 11$

We have an error. Let's go back and check the problem statement, specifically the point $(\alpha, \alpha+1)$. The problem states that the point is $(\alpha, \alpha+1)$, not that it is on the curve.

Let the point on the curve be $(\alpha, f(\alpha))$, where $f(x) = \frac{1}{4}(x+1)^2$. Then, $f'(\alpha) = \frac{1}{2}(\alpha+1)$. The slope of the normal is therefore $-\frac{2}{\alpha+1}$. The equation of the normal at $(\alpha, \frac{1}{4}(\alpha+1)^2)$ is:

$y - \frac{1}{4}(\alpha+1)^2 = -\frac{2}{\alpha+1}(x-\alpha)$

We want to find the x-intercept, so we set $y=0$: $-\frac{1}{4}(\alpha+1)^2 = -\frac{2}{\alpha+1}(x-\alpha)$ $\frac{1}{4}(\alpha+1)^3 = 2(x-\alpha)$ $(\alpha+1)^3 = 8(x-\alpha)$ $x = \alpha + \frac{1}{8}(\alpha+1)^3$

We are given that the x-intercept must be 2. Therefore, $2 = \alpha + \frac{1}{8}(\alpha+1)^3$ $16 = 8\alpha + (\alpha+1)^3$ $16 = 8\alpha + \alpha^3 + 3\alpha^2 + 3\alpha + 1$ $\alpha^3 + 3\alpha^2 + 11\alpha - 15 = 0$

We can see that $\alpha = 1$ is a root: $1 + 3 + 11 - 15 = 0$ So, $(\alpha - 1)$ is a factor. Dividing the cubic by $(\alpha - 1)$ gives: $\alpha^2 + 4\alpha + 15 = 0$ This quadratic has no real roots since $b^2-4ac = 16 - 60 = -44 < 0$. Therefore, $\alpha = 1$.

The derivative at $x=1$ is $f'(1) = \frac{1}{2}(1+1) = 1$. So, the slope of the normal is $-1$. The point on the curve is $(1, f(1)) = (1, \frac{1}{4}(1+1)^2) = (1, 1)$. The equation of the normal is: $y - 1 = -1(x - 1)$ $y - 1 = -x + 1$ $y = -x + 2$

The x-intercept is when $y = 0$: $0 = -x + 2$ $x = 2$

Common Mistakes & Tips

• Careless Algebra: Double-check your algebraic manipulations, especially when solving systems of equations.
• Interpreting Tangency: Remember that tangency implies both a common point and equal slopes.
• Normal vs. Tangent: Always take the negative reciprocal of the tangent's slope to find the normal's slope.

Summary

We first determined the equation of the quadratic curve using the given conditions. Then, we found the derivative to find the slope of the tangent. Next, we calculated the slope of the normal and its equation. Finally, we found the x-intercept of the normal line.

The final answer is \boxed{2}.