Question: f(x) = ax² + 12x + c
In the given quadratic function, a and c are constants. The graph of y = f(x) in the xy plane is a parabola that opens upward and has a vertex at the point (h, k), where h and k are constants. If k < 0 and f(1) = f(-5), which of the following must be true?
I. c < 0
II. a ≥ 1
A I only
B II only
C I and II
D Neither I nor II
1. **State the problem:** We have a quadratic function $$f(x) = ax^2 + 12x + c$$ with constants $$a$$ and $$c$$. The parabola opens upward, so $$a > 0$$. The vertex is at $$(h,k)$$ with $$k < 0$$, and we know $$f(1) = f(-5)$$. We want to determine which statements about $$c$$ and $$a$$ must be true.
2. **Recall vertex formula:** The vertex $$h$$ of a parabola $$y = ax^2 + bx + c$$ is at $$h = -\frac{b}{2a}$$. Here, $$b = 12$$, so
$$
h = -\frac{12}{2a} = -\frac{6}{a}
$$
3. **Calculate vertex value $$k$$:**
$$
k = f(h) = a\left(-\frac{6}{a}\right)^2 + 12\left(-\frac{6}{a}\right) + c = a\frac{36}{a^2} - \frac{72}{a} + c = \frac{36}{a} - \frac{72}{a} + c = -\frac{36}{a} + c
$$
4. **Given $$k < 0$$:**
$$
-\frac{36}{a} + c < 0 \implies c < \frac{36}{a}
$$
Since $$a > 0$$, this inequality relates $$c$$ and $$a$$.
5. **Use condition $$f(1) = f(-5)$$:**
Calculate $$f(1)$$:
$$
f(1) = a(1)^2 + 12(1) + c = a + 12 + c
$$
Calculate $$f(-5)$$:
$$
f(-5) = a(-5)^2 + 12(-5) + c = 25a - 60 + c
$$
Set equal:
$$
a + 12 + c = 25a - 60 + c
$$
Cancel $$c$$ on both sides:
$$
a + 12 = 25a - 60
$$
Rearranged:
$$
12 + 60 = 25a - a \implies 72 = 24a \implies a = 3
$$
6. **Substitute $$a = 3$$ into inequality for $$c$$:**
$$
c < \frac{36}{3} = 12
$$
7. **Analyze statements:**
- Statement I: $$c < 0$$. We only know $$c < 12$$, so $$c$$ could be negative or positive less than 12. So I is **not necessarily true**.
- Statement II: $$a \geq 1$$. We found $$a = 3$$, so II is **true**.
**Final answer:** B II only