Question

Find the roots and the vertex of the quadratic on a calculator. Round all values to 3 decimal places (if necessary).
y, equals, x, squared, plus, 2, x, minus, 120
y=x
2
+2x−120

Answer 1

To find the roots and vertex of the quadratic equation \( y = x^2 + 2x - 120 \), we can use the quadratic formula and vertex formula.

### Step 1: Finding the Roots

The quadratic formula for finding the roots of \( ax^2 + bx + c = 0 \) is given by:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \( y = x^2 + 2x - 120 \):
- \( a = 1 \)
- \( b = 2 \)
- \( c = -120 \)

Now, we calculate the discriminant:
\[
b^2 - 4ac = 2^2 - 4(1)(-120) = 4 + 480 = 484
\]

Now apply the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{484}}{2(1)} = \frac{-2 \pm 22}{2}
\]

Calculating the two potential roots:
1. \( x_1 = \frac{-2 + 22}{2} = \frac{20}{2} = 10 \)
2. \( x_2 = \frac{-2 - 22}{2} = \frac{-24}{2} = -12 \)

So the roots of the quadratic are:
\[
x_1 = 10 \quad \text{and} \quad x_2 = -12
\]

### Step 2: Finding the Vertex

The vertex of a quadratic \( y = ax^2 + bx + c \) can be found using the formula:
\[
x = -\frac{b}{2a}
\]

Substituting the values of \( a \) and \( b \):
\[
x = -\frac{2}{2(1)} = -\frac{2}{2} = -1
\]

Now, to find the y-coordinate of the vertex, substitute \( x = -1 \) back into the original function:
\[
y = (-1)^2 + 2(-1) - 120 = 1 - 2 - 120 = -121
\]

### Final Results

Thus, the roots of the quadratic are:
- \( x_1 = 10 \)
- \( x_2 = -12 \)

The vertex of the quadratic is:
- Vertex: \( (-1, -121) \)

### Summary
- Roots: \( 10 \) and \( -12 \)
- Vertex: \( (-1, -121) \)