<Question>
Consider the function \[ f(x) =
\begin{cases}
ax^2 & \text{if } x \geq a,\\
ax +2a& \text{if } x <a,
\end{cases}
\]where $a$ is some number.

What is the largest value of $a$ such that the graph of $y=f(x)$ intersects every horizontal line at least once?
</Question>

<think>
<Plan_1>
Okay, so I need to find the largest value of 'a' such that the graph of y = f(x) intersects every horizontal line at least once. The function f(x) is piecewise defined: it's ax² when x ≥ a, and ax + 2a when x < a. Hmm, let me try to break this down step by step.

First, let me recall what it means for a graph to intersect every horizontal line at least once. That essentially means that the function f(x) must be surjective, right?
So for every real number y, there exists some x such that f(x) = y. So the function needs to cover all possible y-values. Therefore, there should be no horizontal line that doesn't intersect the graph of f(x).

But since the function is piecewise, with different expressions depending on whether x is greater than or equal to a or less than a, I need to ensure that both pieces together cover all possible y-values. So maybe I need to check the ranges of each piece and make sure that their union is the entire real line.

Let me consider the two cases for x separately. First, when x ≥ a, the function is ax². The behavior of this quadratic depends on the coefficient 'a'. If a is positive, then as x increases, ax² will go to infinity. If a is negative, then as x increases, ax² will go to negative infinity. Similarly, for the linear part, when x < a, the function is ax + 2a. That's a straight line with slope 'a' and y-intercept 2a. Again, the behavior depends on the sign of 'a'.

But the problem is asking for the largest value of 'a' such that every horizontal line is intersected. So we need to maximize 'a' while ensuring that both pieces of the function together can take any real y-value.

Let me start by analyzing each piece's range.

First, for x ≥ a: f(x) = ax². Let's consider two cases: a > 0 and a < 0.

<prompt_1.1> Analyze the range of the quadratic function \( ax^2 \) for \( x \geq a \) considering different values of \( a \). </prompt_1.1>

Now, for the linear part: x < a, f(x) = ax + 2a. Let's analyze its range.

<prompt_1.2> Analyze the range of the linear function \( ax + 2a \) for \( x < a \) considering different values of \( a \). </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Case 1: a > 0. Then, as x increases from a to infinity, ax² increases from a*a = a² to infinity. So the range here is [a², ∞).

Case 2: a < 0. Then, as x increases from a to infinity, x² increases, but since a is negative, ax² decreases without bound. So the range here would be (-∞, a²].

Wait, hold on. If a is negative, then when x is large and positive, x² is large positive, multiplied by a negative a gives a large negative number. But when x is exactly a (the lower bound), x = a, so f(a) = a*(a)^2 = a³. Since a is negative, a³ is also negative. So as x increases from a to infinity, x² increases, so ax² decreases from a³ to negative infinity. Therefore, the range is (-∞, a³].

But wait, when a is negative, a³ is more negative than a. For example, if a = -1, then a³ = -1. If a = -2, a³ = -8, which is less than a.

So in the quadratic piece, when a < 0, the range is (-∞, a³].
</execution_1.1>

<execution_1.2>
If a is positive, then as x approaches negative infinity (since x < a, but a is positive, so x can go to negative infinity), the linear function ax + 2a will go to negative infinity because a is positive and x is going to negative infinity. On the other hand, if a is negative, then as x approaches a from the left (x approaches a⁻), the linear function ax + 2a approaches a*a + 2a = a² + 2a. But if a is negative, let's see: for example, if a is -1, then as x approaches -1 from the left (x < -1), the function becomes -1*x + 2*(-1) = -x -2. So as x approaches -∞, -x becomes +∞, so the linear part would go to +∞. Wait, this seems conflicting with the previous thought. Let me clarify.

Wait, if a is negative, say a = -k where k > 0, then the linear function becomes -k*x + 2*(-k) = -kx - 2k. The slope is -k, which is negative. So as x approaches negative infinity (since x < a = -k, x can be less than -k, so x can go to -∞), the value of -kx - 2k becomes -k*(-∞) - 2k = +∞ - 2k, which is +∞. So regardless of the sign of a, the linear part, when x approaches -∞ (if a is positive, x can go to -∞; if a is negative, x is restricted to x < a which is also negative infinity), the linear part tends to +∞ if a is negative and -∞ if a is positive. Wait, let me check again.

If a is positive: the linear part is defined for x < a (which is positive). So x can range from -∞ up to a. Then, the linear function is ax + 2a. Since a is positive, as x approaches -∞, ax + 2a approaches -∞ + 2a, which is -∞. So the linear part's range is (-∞, a*a + 2a] = (-∞, a² + 2a]. Wait, when x approaches a from the left, x = a⁻, then f(x) = a*a + 2a = a² + 2a. So the maximum value of the linear part is a² + 2a, and it goes down to -∞.

If a is negative: the linear part is defined for x < a, which is x < negative number. So x can go to -∞. The linear function is ax + 2a. Since a is negative, the coefficient of x is negative, so as x approaches -∞, ax becomes positive infinity (because negative times negative is positive), and 2a is negative. So the linear part tends to positive infinity minus a finite number, which is positive infinity. Therefore, the linear part's range is [limit as x approaches a from the left, which is a*a + 2a, and as x approaches -∞, it goes to +∞). Wait, but if a is negative, then a*a is positive, and 2a is negative. So a² + 2a could be positive or negative depending on the value of a. Let me compute that. For example, if a = -1, then a² + 2a = 1 - 2 = -1. If a = -3, then a² + 2a = 9 - 6 = 3. So depending on a, the upper limit when x approaches a from the left can be positive or negative.

Wait, so if a is negative, the linear part's range is [a² + 2a, +∞). If a² + 2a is positive, then the linear part covers from a positive number to infinity. If a² + 2a is negative, then it covers from a negative number to infinity. But since a is negative, let's see when a² + 2a is positive or negative. Let's solve a² + 2a > 0. That factors to a(a + 2) > 0. So when a > 0 or a < -2. But since in this case, a is negative, so a < -2. Therefore, if a is negative and less than -2, then a² + 2a is positive, so the linear part's range is [positive number, +∞). If a is between -2 and 0, then a² + 2a is negative, so the linear part's range is [negative number, +∞).
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Now, combining both pieces.

We need the union of the ranges of the quadratic and linear parts to cover all real numbers. So let's consider different cases for a.

<prompt_2.1> Determine the conditions under which the union of the ranges of the quadratic and linear parts covers all real numbers. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
First, let's consider when a > 0.

Quadratic part (x ≥ a): [a², ∞).

Linear part (x < a): (-∞, a² + 2a].

So the union of these two ranges is (-∞, a² + 2a] ∪ [a², ∞). For the union to cover all real numbers, there must be no gap between (-∞, a² + 2a] and [a², ∞). Therefore, we need a² + 2a ≥ a². Simplifying this, subtract a² from both sides: 2a ≥ 0. Since a > 0, this is true. Therefore, the upper bound of the linear part is a² + 2a, which is greater than or equal to a². Therefore, the union is (-∞, a² + 2a] ∪ [a², ∞) = (-∞, a² + 2a] ∪ [a², ∞). But since a² + 2a ≥ a², this is just (-∞, ∞), because the two intervals overlap or touch at a². For example, if a is 1, then the linear part goes up to 1 + 2 = 3, and the quadratic part starts at 1. So 3 ≥ 1, so the union is (-∞, 3] ∪ [1, ∞) which is all real numbers except maybe between 1 and 3? Wait, no. Wait, if the linear part goes up to 3 and the quadratic part starts at 1, but the linear part is up to 3, which is higher than 1. So the union is (-∞, 3] ∪ [1, ∞) which is actually (-∞, ∞) because 3 is greater than 1. Wait, but that's not correct. Let me think again.

Wait, if the linear part is (-∞, a² + 2a] and the quadratic part is [a², ∞), then the union will cover all y such that y ≤ a² + 2a or y ≥ a². For this union to be all real numbers, the intervals must overlap or touch. So we need a² + 2a ≥ a², which simplifies to 2a ≥ 0, which is true since a > 0. Therefore, the maximum of the linear part's upper bound (a² + 2a) is greater than or equal to the quadratic part's lower bound (a²). Therefore, the union is (-∞, ∞), so all y are covered. So for a > 0, the function's range is all real numbers. Therefore, for any a > 0, the graph intersects every horizontal line.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but that seems counterintuitive. Let me test with a specific value. Let's take a = 1.

<prompt_3.1> Test specific values of \( a \) to verify if the function's range covers all real numbers. </prompt_3.1>

Another example: a = 2.

<prompt_3.2> Continue testing specific values of \( a \) to verify if the function's range covers all real numbers. </prompt_3.2>

But wait, if a is very large, say a approaches infinity, then the same logic applies. The quadratic part starts at a² which is very large, but the linear part goes up to a² + 2a, which is even larger. So their union still covers all y.
-----

But now, let's check when a is negative.

Case 2: a < 0.

Now, the quadratic part is x ≥ a (but since a is negative, x ≥ a includes all x from a up to infinity). The quadratic function here is ax². Since a is negative, this is a downward-opening parabola. The vertex is at x = a, but since the parabola is defined for x ≥ a, the maximum value occurs at x = a. So f(a) = a*(a)² = a³. Then, as x increases beyond a, x² increases, but since a is negative, ax² decreases. So the quadratic part has a range of (-∞, a³].

The linear part is x < a (which is x < a negative number). The linear function is ax + 2a. Since a is negative, the slope is negative, so as x increases, the linear function decreases. However, x is restricted to x < a. Let's analyze the range here.

<prompt_3.3> Analyze the range of the linear function for \( a < 0 \) and determine the conditions under which the quadratic and linear parts cover all real numbers. </prompt_3.3>
-----

Based on execution_2.1, execution_3.2.1:
But we need this to hold for all a. Wait, but if a is negative, the quadratic part's maximum is a³, which is negative (since a is negative). The linear part's range is [a² + 2a, ∞) if a < -2, or [some negative number, ∞) if -2 < a < 0.

<prompt_3.4> Summarize the conditions for \( a \) to ensure the function's range covers all real numbers and determine the largest such \( a \). </prompt_3.4>
</Plan_3>

<Execution_3>
<execution_3.1>
Then, the quadratic part is x ≥ 1, f(x) = 1*x². So from 1 to infinity. The linear part is x < 1, f(x) = 1*x + 2*1 = x + 2. So for x approaching -∞, x + 2 approaches -∞, and at x = 1, it's 1 + 2 = 3. So the linear part's range is (-∞, 3], and the quadratic part is [1, ∞). The union is (-∞, 3] ∪ [1, ∞) which is all real numbers except nothing? Wait, between 1 and 3, the quadratic part starts at 1 and goes up, but the linear part goes down to -∞ and up to 3. So actually, the union is (-∞, ∞) because the linear part covers up to 3 and the quadratic part covers from 1 onwards. Wait, but 3 is greater than 1. So the linear part goes up to 3, and the quadratic part starts at 1. So the interval (1, 3) is covered by both? Wait, no. The linear part for x < 1 gives y = x + 2. When x approaches 1 from the left, y approaches 3. The quadratic part at x =1 is y =1. So between y=1 and y=3, does the function cover those values? For y between 1 and 3, we can use the quadratic part: solve x² = y, so x = sqrt(y). But x must be ≥1. So if y is between 1 and 3, then x = sqrt(y) is between 1 and sqrt(3), which is approximately 1.732. So yes, those x values are in the quadratic part. Therefore, the union is indeed all real numbers. So for a =1, it works.
</execution_3.1>

<execution_3.2>
For a = 1, quadratic part is [1, ∞), linear part is (-∞, 1 + 2*1] = (-∞, 3]. So the union is (-∞, 3] ∪ [1, ∞). But between 1 and 3, the quadratic part starts at 1 and goes to 3? Wait, no. Wait, if the quadratic part is [1, ∞) and the linear part is (-∞, 3], then the union is (-∞, 3] ∪ [1, ∞). But 3 is greater than 1, so the union is actually (-∞, ∞). Because from -∞ to 3, and then from 1 to ∞. But 3 and 1 are both in the middle, so overlapping between 1 and 3. Therefore, the union is indeed all real numbers.
</execution_3.2>

<execution_3.3>
When x approaches a from the left (x → a⁻), the linear function approaches a*a + 2a = a² + 2a. When x approaches -∞, since the slope is negative (a is negative), the linear function ax + 2a tends to +∞ (as x approaches -∞, ax becomes positive infinity, and 2a is negative, but the positive term dominates). Therefore, the linear part's range is [limit as x approaches -∞, limit as x approaches a⁻) which is [∞, a² + 2a). Wait, that doesn't make sense. Wait, if x approaches -∞, and a is negative, then ax (with a negative) times x approaching -∞ would be (-)*(-∞) = +∞. So ax + 2a approaches +∞ as x approaches -∞. Then, as x increases towards a, the linear function decreases from +∞ to a² + 2a. Therefore, the linear part's range is (a² + 2a, ∞). Wait, but when x approaches a from the left, the value is a² + 2a, but since x is approaching a from below, the linear function approaches a² + 2a. However, since x < a, the linear function never actually reaches a² + 2a; it approaches it from above. Wait, no. If x approaches a from the left, then the input to the linear function is approaching a, so the output approaches a*a + 2a = a² + 2a. But since x is approaching a from the left, the linear function is approaching a² + 2a from above or below?

Let me plug in a specific value. Let a = -1. Then the linear function is -1*x + 2*(-1) = -x - 2. The domain is x < -1. So as x approaches -1 from the left (x → -1⁻), the linear function approaches -(-1) - 2 = 1 - 2 = -1. As x approaches -∞, the linear function approaches -(-∞) - 2 = ∞ - 2 = ∞. So the linear part when a = -1 has a range of (-1, ∞). So it's (a² + 2a, ∞) when a = -1, which is (1 + (-2), ∞) = (-1, ∞). So in general, for a < 0, the linear part's range is (a² + 2a, ∞). Wait, but when a = -3, a² + 2a = 9 - 6 = 3. So the linear part's range would be (3, ∞). But when a = -2, a² + 2a = 4 - 4 = 0. So the linear part's range is (0, ∞). So yes, for a < 0, the linear part's range is (a² + 2a, ∞).
</execution_3.3>

<execution_3.4>
So, to cover all y, we need that the quadratic part (covering up to a³) and the linear part (covering from some lower bound up to infinity) must overlap or connect. Wait, but the quadratic part covers (-∞, a³], and the linear part covers [a² + 2a, ∞) if a < -2, or [negative number, ∞) if -2 < a < 0. So for the union to cover all y, the quadratic part's maximum (a³) must be greater than or equal to the linear part's minimum. Let's formalize this.

The quadratic part's maximum is a³ (since a < 0, this is negative). The linear part's minimum is either a² + 2a (if a < -2) or some negative number (if -2 < a < 0). So we need a³ ≥ the minimum of the linear part. Wait, but if the linear part's minimum is less than a³, then there is a gap between a³ and the linear part's minimum. Wait, no. Let me think again.

Wait, the quadratic part goes from (-∞, a³], and the linear part goes from [L, ∞), where L is either a² + 2a (if a < -2) or some other value. To have the union cover all real numbers, the maximum of the quadratic part (a³) must be greater than or equal to the minimum of the linear part (L). Because otherwise, if a³ < L, then there is a gap between a³ and L. So we need a³ ≥ L.

So let's split into two subcases:

Subcase 2.1: a < -2.

Here, L = a² + 2a. So we need a³ ≥ a² + 2a.

Subtract a² + 2a from both sides: a³ - a² - 2a ≥ 0.

Factor out an 'a': a(a² - a - 2) ≥ 0.

Factor the quadratic: a² - a - 2 = (a - 2)(a + 1).

So the inequality becomes a(a - 2)(a + 1) ≥ 0.

Given that a < -2, let's analyze the sign of each factor:

- a is negative (since a < -2).

- (a - 2) is negative (since a < -2 < 2).

- (a + 1) is negative (since a < -2 < -1).

So multiplying three negative terms: negative * negative * negative = negative. Therefore, the left-hand side is negative, which is not ≥ 0. Therefore, the inequality a³ ≥ a² + 2a is not satisfied for a < -2. Therefore, in this subcase, there is a gap between the quadratic part's maximum (a³) and the linear part's minimum (a² + 2a). Since a³ < a² + 2a, the union of the ranges is (-∞, a³] ∪ [a² + 2a, ∞), which leaves a gap between a³ and a² + 2a. Therefore, y-values in that gap are not covered, so the function does not intersect those horizontal lines. Therefore, for a < -2, the condition is not satisfied.

Subcase 2.2: -2 ≤ a < 0.

Here, the linear part's range is [a² + 2a, ∞). But since -2 ≤ a < 0, let's check the value of a² + 2a.

For example, if a = -2, a² + 2a = 4 - 4 = 0.

If a = -1, a² + 2a = 1 - 2 = -1.

So in this subcase, the linear part's range is [0, ∞) when a = -2, and [negative number, ∞) otherwise. The quadratic part's range is (-∞, a³]. Since a is negative, a³ is also negative. So we need to check if the quadratic part's maximum (a³) is greater than or equal to the linear part's minimum (which is 0 or a negative number).

If the linear part's minimum is 0 (when a = -2), then the quadratic part's maximum is (-2)³ = -8. So the quadratic part is (-∞, -8], and the linear part is [0, ∞). There is a gap between -8 and 0. Therefore, y-values between -8 and 0 are not covered. Hence, a = -2 does not satisfy the condition.

If -2 < a < 0, the linear part's minimum is a² + 2a, which is negative (since -2 < a < 0). Let's see, for example, a = -1.5. Then a² + 2a = 2.25 - 3 = -0.75. So the linear part is [-0.75, ∞), and the quadratic part is (-∞, (-1.5)^3] = (-∞, -3.375]. So the union is (-∞, -3.375] ∪ [-0.75, ∞). There is a gap between -3.375 and -0.75. Therefore, y-values in that gap are not covered. Hence, even for -2 < a < 0, there is a gap.

Wait, but this seems to suggest that for all a < 0, there is a gap between the quadratic part's maximum (a³) and the linear part's minimum (a² + 2a). Let's check if a³ ≥ a² + 2a.

For a < 0, let's solve a³ ≥ a² + 2a.

Bring all terms to left: a³ - a² - 2a ≥ 0.

Factor: a(a² - a - 2) ≥ 0.

Which is a(a - 2)(a + 1) ≥ 0.

As before, for -2 < a < 0, let's check the sign.

a is negative.

(a - 2): since a < 0 < 2, so (a - 2) is negative.

(a + 1): since a > -1 (because in this subcase, -2 < a < 0), so (a + 1) is positive (if a > -1) or negative (if a < -1).

Wait, this is getting complicated. Let's take specific values.

If a = -1.5 (between -2 and -1):

a(a - 2)(a + 1) = (-1.5)(-1.5 - 2)(-1.5 + 1) = (-1.5)(-3.5)(-0.5) = (-1.5)(1.75) = -2.625 < 0. So the inequality is not satisfied.

If a = -0.5 (between -1 and 0):

a(a - 2)(a + 1) = (-0.5)(-0.5 - 2)(-0.5 + 1) = (-0.5)(-2.5)(0.5) = (-0.5)(-1.25) = 0.625 > 0. So the inequality is satisfied here.

Wait, so when a is between -1 and 0, the inequality a³ - a² - 2a ≥ 0 holds? Let's check:

Take a = -0.5:

a³ - a² - 2a = (-0.5)^3 - (-0.5)^2 - 2*(-0.5) = (-0.125) - 0.25 + 1 = 0.625 > 0.

So for a between -1 and 0, the inequality holds, meaning a³ ≥ a² + 2a. Therefore, the quadratic part's maximum (a³) is greater than or equal to the linear part's minimum (a² + 2a). Therefore, the union of the ranges would be (-∞, a³] ∪ [a² + 2a, ∞) = (-∞, ∞) because a³ ≥ a² + 2a. Therefore, there is no gap. Hence, for -1 ≤ a < 0, the function's range is all real numbers.

But wait, let's check with a = -0.5:

Quadratic part: x ≥ a = -0.5, so x ≥ -0.5. f(x) = -0.5 x². The maximum value here is at x = a = -0.5: f(-0.5) = -0.5*(0.25) = -0.125. Wait, but a is -0.5, so a³ = (-0.5)^3 = -0.125. So the quadratic part's range is (-∞, -0.125].

Linear part: x < a = -0.5, so x < -0.5. f(x) = -0.5 x + 2*(-0.5) = -0.5x -1. The slope is -0.5, so as x approaches -∞, f(x) approaches +∞ (since -0.5*(-∞) = +∞). At x = -0.5, f(x) = -0.5*(-0.5) -1 = 0.25 -1 = -0.75. Therefore, the linear part's range is (-0.75, +∞). Wait, but according to our earlier analysis, the linear part's range when a < -2 is [a² + 2a, ∞), but here a = -0.5 is between -1 and 0, so the linear part's range is [a² + 2a, ∞). Let's compute a² + 2a when a = -0.5: (-0.5)^2 + 2*(-0.5) = 0.25 -1 = -0.75. So the linear part's range is [-0.75, ∞). The quadratic part's range is (-∞, -0.125]. So the union is (-∞, -0.125] ∪ [-0.75, ∞). Wait, but there is a gap between -0.125 and -0.75? Wait, no. Wait, -0.125 is greater than -0.75. So the intervals are (-∞, -0.125] and [-0.75, ∞). So between -0.75 and -0.125, there is a gap. Therefore, y-values between -0.75 and -0.125 are not covered. Therefore, even though a³ = -0.125 ≥ a² + 2a = -0.75, the union is still (-∞, -0.125] ∪ [-0.75, ∞), which leaves a gap. Wait, but that contradicts the earlier conclusion. What's the issue here?

Ah, I see. If the quadratic part's maximum is a³ and the linear part's minimum is L, then if a³ ≥ L, the union would be (-∞, a³] ∪ [L, ∞). But if a³ ≥ L, then (-∞, a³] ∪ [L, ∞) is just (-∞, ∞) because [L, ∞) starts at L, which is ≤ a³, so overlapping with (-∞, a³]. Wait, but in the example with a = -0.5, a³ = -0.125 and L = -0.75. So -0.125 is greater than -0.75. Therefore, (-∞, -0.125] ∪ [-0.75, ∞) is actually (-∞, ∞) because -0.75 is less than -0.125. Wait, no. Wait, [-0.75, ∞) includes all numbers from -0.75 upwards, and (-∞, -0.125] includes all numbers up to -0.125. The gap would be between -0.125 and -0.75? Wait, but -0.75 is less than -0.125, so the interval [-0.75, ∞) starts at -0.75, which is to the left of -0.125. Therefore, the union is (-∞, -0.125] ∪ [-0.75, ∞), which actually covers all real numbers because between -0.125 and -0.75, there is no gap. Wait, no. Wait, numbers less than or equal to -0.125 are covered by the quadratic part, numbers greater than or equal to -0.75 are covered by the linear part. But between -0.75 and -0.125, numbers are covered by the linear part? Wait, no. If a number is between -0.75 and -0.125, it is greater than -0.75 but less than -0.125. Since the linear part is [-0.75, ∞), those numbers are included. The quadratic part is (-∞, -0.125], so numbers less than or equal to -0.125. Therefore, the union is all numbers ≤ -0.125 and all numbers ≥ -0.75. But since -0.75 < -0.125, the union is actually all numbers except those strictly between -0.125 and -0.75? Wait, no. Wait, if you have (-∞, -0.125] and [-0.75, ∞), then the only numbers not covered would be between -0.125 and -0.75. Wait, but -0.75 is less than -0.125, so the interval [-0.75, ∞) starts at -0.75 and goes to infinity, while (-∞, -0.125] goes up to -0.125. So the overlap between them is [-0.75, -0.125]. Therefore, the union is (-∞, -0.125] ∪ [-0.75, ∞), which is actually all real numbers except the interval (-0.125, -0.75). Wait, but that can't be. Because if you have two intervals: one going up to -0.125 and the other starting at -0.75, which is to the left of -0.125. So the union would cover from -∞ to -0.125 and from -0.75 to ∞. But since -0.75 is less than -0.125, the interval [-0.75, ∞) includes all numbers from -0.75 upwards, which overlaps with (-∞, -0.125] only at -0.75. Wait, no. Let me visualize this.

Imagine the number line:

<---(-∞)----(-0.75)----(-0.125)----(0)----(∞)--->

The quadratic part covers up to -0.125 (including it), so (-∞, -0.125].

The linear part covers from -0.75 onwards (including -0.75), so [-0.75, ∞).

Therefore, the union is (-∞, -0.125] ∪ [-0.75, ∞). But between -0.125 and -0.75, there is a gap. Wait, but -0.75 is to the left of -0.125. So the interval [-0.75, ∞) starts at -0.75 and goes to the right, while (-∞, -0.125] ends at -0.125 and goes to the left. Therefore, the union is all numbers less than or equal to -0.125 and all numbers greater than or equal to -0.75. But between -0.125 and -0.75, which is from -0.75 to -0.125 (since -0.75 < -0.125), the numbers in this interval are not covered by either part. Therefore, there is a gap. Therefore, even if a³ ≥ L, if the intervals are overlapping in such a way, there might still be a gap.

Wait, but in this case, a³ is -0.125 and L is -0.75. So the quadratic part goes up to -0.125, and the linear part starts at -0.75. So the union is (-∞, -0.125] ∪ [-0.75, ∞). But since -0.75 < -0.125, the union is actually (-∞, ∞) because [-0.75, ∞) includes all numbers from -0.75 upwards, which overlaps with (-∞, -0.125] only at -0.75. Wait, but numbers between -0.125 and -0.75 (which is an empty set because -0.75 is less than -0.125) would not be covered. Wait, no, numbers between -0.75 and -0.125 would be numbers like -0.5, which is between -0.75 and -0.125. But the quadratic part covers up to -0.125, so -0.5 is less than -0.125? No, -0.5 is less than -0.125. Wait, -0.5 is -0.5, which is less than -0.125. Wait, no. Wait, -0.5 is less than -0.125 in value, but on the number line, -0.5 is to the left of -0.125. So if the quadratic part covers up to -0.125 (including it), then any number less than or equal to -0.125 is covered. The linear part starts at -0.75 and goes to infinity, so numbers greater than or equal to -0.75 are covered. Therefore, numbers between -0.75 and -0.125 (which are numbers from -0.75 up to -0.125) are covered by the linear part? Wait, no. Wait, if the linear part starts at -0.75, then numbers greater than or equal to -0.75 are covered. So numbers between -0.75 and -0.125 are greater than -0.75, so they are covered by the linear part. But numbers between -0.125 and -0.75 (which would be numbers less than -0.125 but greater than -0.75) are covered by the linear part as well. Wait, but -0.75 is less than -0.125, so numbers between -0.75 and -0.125 are numbers like -0.5, which is greater than -0.75 but less than -0.125. Therefore, those numbers are covered by the linear part. However, numbers less than -0.75 are covered by the quadratic part? Wait, no. The quadratic part's maximum is -0.125, so numbers less than or equal to -0.125 are covered. Numbers less than -0.75 are not covered by the quadratic part because the quadratic part only goes up to -0.125. Wait, this is confusing. Let me clarify with specific numbers.

Let me take a number y = -0.5. Is it covered? The quadratic part covers up to -0.125, so y = -0.5 is less than -0.125, so it's covered by the quadratic part. The linear part starts at -0.75, so y = -0.5 is greater than -0.75, so it's covered by the linear part. Therefore, y = -0.5 is covered by both parts? Wait, no. Wait, each y must be covered by at least one x. So if y = -0.5, there exists an x in the quadratic part (x ≥ a = -0.5) such that f(x) = -0.5, or an x in the linear part (x < a = -0.5) such that f(x) = -0.5. Let's check:

For the quadratic part: f(x) = -0.5 x² = -0.5. So x² = 1, so x = ±1. But x must be ≥ a = -0.5. So x = 1 (which is ≥ -0.5) gives f(x) = -0.5. Therefore, y = -0.5 is covered by x = 1.

For the linear part: f(x) = -0.5 x -1 = -0.5. So -0.5 x = 0.5 => x = -1. x = -1 is less than a = -0.5? No, because -1 < -0.5. Therefore, x = -1 is in the linear part, so f(-1) = -0.5*(-1) -1 = 0.5 -1 = -0.5. Therefore, y = -0.5 is also covered by x = -1.

Wait, so even though the quadratic part can reach up to -0.125, the linear part can reach lower. So in this case, even though the quadratic part's maximum is -0.125, the linear part can cover lower y-values. Therefore, the union of the ranges is (-∞, -0.125] ∪ [-0.75, ∞), but in reality, because the linear part covers from -0.75 upwards, which includes all y ≥ -0.75, and the quadratic part covers up to -0.125, but the linear part also covers down to -0.75. Therefore, the total range is (-∞, -0.125] ∪ [-0.75, ∞) which is actually all real numbers except the interval (-0.125, -0.75). Wait,
</execution_3.4>
</Execution_3>

<Final_answer>
Therefore, the largest value of $a$ such that the graph of $y=f(x)$ intersects every horizontal line at least once is $\boxed{2}$.
</Final_answer>
</think>