Question

y = x^2 + 2x - 8

Answer 1

The equation \( y = x^2 + 2x - 8 \) represents a quadratic function. We can analyze it further by finding its important features such as the vertex, axis of symmetry, and x-intercepts.

### 1. Standard Form:
The equation is already in the standard form of a quadratic function:
\[ y = ax^2 + bx + c \]
where \( a = 1 \), \( b = 2 \), and \( c = -8 \).

### 2. Finding the Vertex:
The x-coordinate of the vertex for a quadratic function in standard form is given by the formula:
\[ x = -\frac{b}{2a} \]

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

Now, substitute \( x = -1 \) back into the function to find the y-coordinate:
\[ y = (-1)^2 + 2(-1) - 8 \]
\[ y = 1 - 2 - 8 = -9 \]

Thus, the vertex is at the point \( (-1, -9) \).

### 3. Axis of Symmetry:
The axis of symmetry is the vertical line that passes through the vertex. Therefore, the equation of the axis of symmetry is:
\[ x = -1 \]

### 4. Finding the x-intercepts:
To find the x-intercepts, set \( y \) to 0 and solve for \( x \):
\[ 0 = x^2 + 2x - 8 \]

We can factor this quadratic equation:
\[ 0 = (x + 4)(x - 2) \]

Setting each factor to zero gives:
1. \( x + 4 = 0 \) → \( x = -4 \)
2. \( x - 2 = 0 \) → \( x = 2 \)

So, the x-intercepts are at the points \( (-4, 0) \) and \( (2, 0) \).

### 5. Finding the y-intercept:
To find the y-intercept, set \( x = 0 \):
\[ y = (0)^2 + 2(0) - 8 \]
\[ y = -8 \]

Thus, the y-intercept is at the point \( (0, -8) \).

### Summary:
- Vertex: \( (-1, -9) \)
- Axis of Symmetry: \( x = -1 \)
- X-intercepts: \( (-4, 0) \) and \( (2, 0) \)
- Y-intercept: \( (0, -8) \)

This information gives a comprehensive overview of the quadratic function \( y = x^2 + 2x - 8 \).