<Question>
The graph of $f(x)=\frac{2x}{x^2-5x-14}$ has vertical asymptotes $x=a$ and $x=b$, and horizontal asymptote $y=c$.  Find $a+b+c$.
</Question>

<think>
<Plan_1>
Okay, so I need to find the vertical and horizontal asymptotes of the function f(x) = (2x)/(x² -5x -14), and then add their x-values (for the vertical asymptotes) and the y-value (for the horizontal asymptote) together. Let me start by recalling what asymptotes are. 

Vertical asymptotes occur where the denominator is zero, but the numerator isn't zero at those points. That's where the function grows without bound. For horizontal asymptotes, it depends on the degrees of the numerator and the denominator. If the degree of the numerator is less than the denominator, the horizontal asymptote is y=0. If they're equal, it's the ratio of the leading coefficients. If the numerator's degree is higher, there's no horizontal asymptote, but maybe an oblique one. 

First, let's tackle the vertical asymptotes. The denominator here is x² -5x -14. To find where that's zero, I need to factor it or use the quadratic formula. Let me try factoring first. Let's see, I need two numbers that multiply to -14 and add to -5.

<prompt_1.1> Factor the denominator x² - 5x - 14 to find the vertical asymptotes. </prompt_1.1>
-----

Now, moving on to the horizontal asymptote. The degrees of the numerator and denominator are 1 and 2, respectively. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote is y=0. So c=0.

Therefore, adding them together: a + b + c = 7 + 0 = 7.
-----

But just to be thorough, let me consider if there's any possibility that the function could have a horizontal asymptote at a different value. For example, sometimes if the degrees are equal, you take the ratio of leading coefficients.

<prompt_1.2> Double-check the conditions for horizontal asymptotes and conclude the solution. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
Hmm. Let's think:  Factors of -14 could be  -7 and 2. Because -7 * 2 = -14, and -7 + 2 = -5. Yes! So the denominator factors as (x - 7)(x + 2). Let me check that: (x -7)(x +2) = x² +2x -7x -14 = x² -5x -14. Perfect. 

So the denominator is zero when x -7 = 0 or x +2 = 0, which is x=7 and x=-2. Therefore, the vertical asymptotes are at x=7 and x=-2. So a and b are 7 and -2.
</execution_1.1>

<execution_1.2>
But here, the numerator's degree is less, so the horizontal asymptote is y=0. Also, let me confirm the factorization again. The denominator factors into (x-7)(x+2), which gives roots at 7 and -2. Correct. So vertical asymptotes at x=7 and x=-2, sum is 5. Wait, hold on. Wait, a and b are the vertical asymptotes, which are x=7 and x=-2. So a and b are 7 and -2. Then a + b is 7 + (-2) = 5. Then c is 0, so total is 5 + 0 = 5? Wait, wait, wait. Wait, hold on. Wait, the vertical asymptotes are at x = a and x = b. So a and b are 7 and -2. So a + b is 7 + (-2) = 5. Then c is 0, so a + b + c = 5 + 0 = 5.
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, the problem says the vertical asymptotes are x=a and x=b. So, do I need to specify both values? The question is asking for a + b + c. So I need to add the two vertical asymptote values and the horizontal asymptote value.
So, a and b are 7 and -2. Let me confirm that. Yes, vertical asymptotes occur where the denominator is zero, so x=7 and x=-2. Then the horizontal asymptote is y=0, so c=0.

Therefore, a + b + c = 7 + (-2) + 0. Let me compute that.
7 plus (-2) is 5, and 5 plus 0 is 5. So the answer should be 5.

Wait, but let me double-check all steps to make sure I didn't make a mistake.

First, factor the denominator.

<prompt_2.1> Verify the factorization of the denominator and the calculation of a + b + c. </prompt_2.1>
-----

Is there any possibility that I made a mistake in adding 7 and -2? 7 + (-2) is 5.
7 - 2 is 5. Correct. And c is definitely 0. So 5 + 0 is 5.

I think that's solid.
I don't see any errors in the reasoning. So the final answer should be 5.

**Final Answer**
\boxed{5}
</Plan_2>

<Execution_2>
<execution_2.1>
Looking for two numbers that multiply to -14 and add to -5. -7 and 2 work. So yes, denominator factors correctly. Then roots at x=7 and x=-2. So vertical asymptotes at those x-values. Then a=7, b=-2. Sum is 5. Horizontal asymptote is y=0 because degree of numerator (1) < degree of denominator (2). So c=0. Total sum 5 + 0 = 5.
</execution_2.1>
</Execution_2>

<Final_answer>
Therefore, the sum of the values of $a$, $b$, and $c$ is $7 + (-2) + 0 = \boxed{5}$.
</Final_answer>
</think>