Let the quadratic curve passing through the point $(-1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f(x)$. Then, the $x$-intercept of the normal to the curve at the point $(\alpha, \alpha + 1)$ in the first quadrant is

1. Setting up the quadratic

Take the most general quadratic

Conditions given:

1. Passes through $(-1,0)$ \Rightarrow $a(-1)^2 + b(-1) + c = 0$ \Rightarrow $a - b + c = 0$.
2. Passes through $(1,1)$ \Rightarrow $a(1)^2 + b(1) + c = 1$ \Rightarrow $a + b + c = 1$.
3. It touches the line $y = x$ at $(1,1)$. "Touch" means the parabola and the line have the same slope there.
   • Slope of $y = x$ is $1$.
   • Derivative of the parabola is $y' = 2ax + b$.
   • At $x = 1$: $2a(1) + b = 1$.

This gives three linear equations which we solve for $a, b, c$.

2. Solving those equations

Write them neatly:

From $2a + b = 1$ we get $b = 1 - 2a$. Plug it into the second equation: $a + (1-2a) + c = 1 \Rightarrow -a + c = 0 \Rightarrow c = a$. Use the first: $a - (1-2a) + c = 0 \Rightarrow 3a - 1 + c = 0$; since $c=a$, $4a = 1 \Rightarrow a = \frac14$. Hence $b = \frac12$, $c = \frac14$.

So the curve is

3. Find the special point $(\alpha, \alpha+1)$ on the curve

Impose

Multiply by $4$:

Either $\alpha+1 = 0 \;(\alpha=-1)$ or $\alpha+1 = 4$.
$\alpha=-1$ is not in the first quadrant, so take $\alpha = 3$. Thus the point is $(3,4)$.

4. Equation of the normal at $(3,4)$

Slope of tangent: $y' = \frac{x+1}{2}$. At $x=3$: $m_\text{tangent} = \frac{3+1}{2} = 2$. Slope of normal:

Normal passing through $(3,4)$:

5. x-intercept (set $y=0$)

Multiply by $-2$:

Hence the normal meets the x-axis at $\boxed{\;x = 11\;}$.

Imagine drawing a smiley bowl–shaped curve (a parabola).

1. The bowl must go through the point (-1,0).
2. It must just kiss (touch) the straight line y = x exactly at the point (1,1); think of two lines lightly touching without crossing.
3. We ask: If we stand somewhere on that bowl where the spot’s addresses look like (number, number+1) and lie in the first quadrant, then drop a stick straight up and down (a normal), where does that stick hit the x-axis?

The steps:

• Build the bowl (find its formula).
• Find the special spot (α, α+1) on the bowl that sits in the first quadrant.
• Write the normal line at that spot.
• See where this line meets the floor (x-axis).

Step-by-step solution

1. Write general quadratic: $y = ax^2 + bx + c$.
2. Use $(-1,0)$ \Rightarrow $a - b + c = 0$.
3. Use $(1,1)$ \Rightarrow $a + b + c = 1$.
4. Tangency at $(1,1)$: slope of parabola $y' = 2ax + b$. At $x=1$: $2a + b = 1$.
5. Solve the simultaneous equations:
   • $b = 1 - 2a$.
   • $c = a$.
   • $4a = 1 \Rightarrow a = \dfrac14$, $b = \dfrac12$, $c = \dfrac14$.
6. Equation of curve: $y = \dfrac{(x+1)^2}{4}$.
7. Let $(\alpha, \alpha+1)$ be on the curve: substitute $x=\alpha$, $y=\alpha+1$: $\dfrac{(\alpha+1)^2}{4} = \alpha + 1.$ Removing common factor $(\alpha+1)$: $\alpha+1 = 4 \;\Rightarrow\; \alpha = 3 \;(\text{first quadrant}).$
8. Point of interest: $(3,4)$.
9. Slope of tangent at $x=3$: $m_t = \dfrac{3+1}{2} = 2$.
10. Slope of normal: $m_n = -\dfrac{1}{2}$.
11. Equation of normal through $(3,4)$: $y - 4 = -\frac12\,(x - 3).$
12. For x-intercept, set $y=0$: $-4 = -\frac12\,(x - 3) \;\Rightarrow\; 8 = x - 3 \;\Rightarrow\; x = 11.$

Answer: $x$-intercept = 11.