Question Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necescary). $$ y=3 x^{2}+9 x-293 $$ Answer Attemptroutof2 Roots: $$ \square $$ and $$ \square $$

Question

UpStudy AI · Accepted Answer

To find the **roots** and the **vertex** of the quadratic function $$ y = 3x^2 + 9x - 293 $$, follow these steps:

### 1. Finding the Roots

The roots of the quadratic equation $$ 3x^2 + 9x - 293 = 0 $$ can be found using the **quadratic formula**:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$ a = 3 $$, $$ b = 9 $$, and $$ c = -293 $$.

**Step-by-Step Calculation:**

1. **Calculate the Discriminant ($$ D $$):**
   $$
   D = b^2 - 4ac = 9^2 - 4(3)(-293) = 81 + 3516 = 3597
   $$
   
2. **Compute the Square Root of the Discriminant:**
   $$
   \sqrt{3597} \approx 59.975
   $$
   
3. **Find the Roots:**
   $$
   x = \frac{-9 \pm 59.975}{2 \times 3} = \frac{-9 \pm 59.975}{6}
   $$
   
   - **First Root:**
     $$
     x_1 = \frac{-9 + 59.975}{6} = \frac{50.975}{6} \approx 8.496
     $$
     
   - **Second Root:**
     $$
     x_2 = \frac{-9 - 59.975}{6} = \frac{-68.975}{6} \approx -11.496
     $$

### 2. Finding the Vertex

The vertex of a quadratic function $$ y = ax^2 + bx + c $$ is given by:
$$
\left( -\frac{b}{2a}, \ y\left(-\frac{b}{2a}\right) \right)
$$

**Step-by-Step Calculation:**

1. **Calculate the x-coordinate of the Vertex:**
   $$
   x_v = -\frac{9}{2 \times 3} = -\frac{9}{6} = -1.5
   $$
   
2. **Calculate the y-coordinate by Substituting $$ x_v $$ into the Original Equation:**
   $$
   y_v = 3(-1.5)^2 + 9(-1.5) - 293 = 3(2.25) - 13.5 - 293 = 6.75 - 13.5 - 293 = -299.75
   $$
   
   $$
   y_v \approx -299.750
   $$

### **Final Answers:**

- **Roots:** $$ 8.496 $$ and $$ -11.496 $$
- **Vertex:** $$ (-1.500, -299.750) $$
Roots: 8.496 and -11.496 Vertex: (-1.500, -299.750)

Question Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necescary). \[ y=3 x^{2}+9 x-293 \] Answer Attemptroutof2 Roots: \( \square \) and \( \square \)

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions