Quiz 2.3 \& 2.4 Question 3 of 10 This quiz: 100 This question: tion list estion 1 lestion 2 uestion 3 uestion 4 question 5 Question 6 Question 7 Question 8 Find the vertex form of the function. Then find each of the following. (A) Intercepts (B) Vertex (C) Maximum or minimum (D) Range $$ s(x)=-3 x^{2}-24 x-43 $$ $$ s(x)= $$ $$ \square $$ (Type your answer in vertex form.) (A) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The $$ y $$-intercept is $$ \square $$ 3. (Type an integer or decimal rounded to two decimal places as needed.) B. There is no $$ y $$-intercept. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The $$ x $$-intercepts are $$ \square $$ . (Use a comma to separate answers as needed. Type an integer or decimal rounded to two decimal places as needed.) B. There is no $$ x $$-intercept. (B) Vertex $$ \square $$ (Type an ordered pair.) (C) The function has a minimum maximum Maximum or minimum value: $$ \square $$ (D) Range: $$ \square $$ (Type your answer as an inequality, or using interval notation.)

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UpStudy AI · Accepted Answer

Let's analyze the quadratic function $$ s(x) = -3x^2 - 24x - 43 $$ by converting it to vertex form and then finding the requested properties.

### **Vertex Form**

To convert $$ s(x) = -3x^2 - 24x - 43 $$ to vertex form $$ s(x) = a(x - h)^2 + k $$, follow these steps:

1. **Factor out the coefficient of $$ x^2 $$:**
   $$
   s(x) = -3(x^2 + 8x) - 43
   $$
2. **Complete the square inside the parentheses:**
   - Take half of the coefficient of $$ x $$ (which is 8), square it, and add and subtract this value inside the parentheses:
     $$
     s(x) = -3\left(x^2 + 8x + 16 - 16\right) - 43 = -3\left((x + 4)^2 - 16\right) - 43
     $$
3. **Simplify the expression:**
   $$
   s(x) = -3(x + 4)^2 + 48 - 43 = -3(x + 4)^2 + 5
   $$
   
**Vertex Form:**
$$
s(x) = -3(x + 4)^2 + 5
$$

---

### **(A) Intercepts**

**1. Y-Intercept:**
To find the y-intercept, evaluate $$ s(x) $$ at $$ x = 0 $$:
$$
s(0) = -3(0)^2 - 24(0) - 43 = -43
$$

**Answer:**
- **A.** The $$ y $$-intercept is $$ -43 $$.

**2. X-Intercepts:**
To find the x-intercepts, set $$ s(x) = 0 $$ and solve for $$ x $$:
$$
-3x^2 - 24x - 43 = 0
$$
Divide the entire equation by -1:
$$
3x^2 + 24x + 43 = 0
$$
Use the quadratic formula:
$$
x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 3 \cdot 43}}{2 \cdot 3} = \frac{-24 \pm \sqrt{576 - 516}}{6} = \frac{-24 \pm \sqrt{60}}{6} = \frac{-24 \pm 2\sqrt{15}}{6} = \frac{-12 \pm \sqrt{15}}{3}
$$
Approximate the solutions:
$$
\sqrt{15} \approx 3.87298
$$
$$
x \approx \frac{-12 + 3.87298}{3} \approx -2.71 \quad \text{and} \quad x \approx \frac{-12 - 3.87298}{3} \approx -5.29
$$

**Answer:**
- **A.** The $$ x $$-intercepts are approximately $$ -2.71 $$ and $$ -5.29 $$.

---

### **(B) Vertex**

From the vertex form $$ s(x) = -3(x + 4)^2 + 5 $$, the vertex $$(h, k)$$ is:

$$
\text{Vertex} = (-4, 5)
$$

**Answer:**
- **B.** Vertex $$ (-4, 5) $$.

---

### **(C) Maximum or Minimum**

Since the coefficient of $$ (x + 4)^2 $$ is $$ -3 $$ (which is less than 0), the parabola opens downward. Therefore, the function has a **maximum** value.

The maximum value is the $$ k $$-value from the vertex form:
$$
\text{Maximum value} = 5
$$

**Answer:**
- **C.** The function has a **maximum**.
- **Maximum value:** $$ 5 $$.

---

### **(D) Range**

Because the parabola opens downward and the maximum value is $$ 5 $$, the range of the function includes all real numbers less than or equal to 5.

**Answer:**
- **D.** Range: $$ s(x) \leq 5 $$ or in interval notation, $$ (-\infty, 5] $$.

---

### **Final Summary**

- **Vertex Form:** $$ s(x) = -3(x + 4)^2 + 5 $$
- **(A) Y-Intercept:** $$ -43 $$
- **(A) X-Intercepts:** $$ -2.71, -5.29 $$
- **(B) Vertex:** $$ (-4, 5) $$
- **(C) Maximum:** $$ 5 $$
- **(D) Range:** $$ (-\infty, 5] $$
**Vertex Form:** $$ s(x) = -3(x + 4)^2 + 5 $$ **(A) Y-Intercept:** $$ -43 $$ **(A) X-Intercepts:** $$ -2.71, -5.29 $$ **(B) Vertex:** $$ (-4, 5) $$ **(C) Maximum:** $$ 5 $$ **(D) Range:** $$ s(x) \leq 5 $$

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