To analyze the quadratic equation y = x² + 6x + 5, we need to find the vertex, axis of symmetry, and intercepts.
1. Vertex
The vertex of a quadratic function in the standard form y = ax² + bx + c can be found using the formula:
x = -b / (2a)
In our equation, a = 1 and b = 6.
Plugging in the values:
x = -6 / (2 * 1) = -6 / 2 = -3
To find the y-coordinate of the vertex, substitute x = -3 back into the original equation:
y = (-3)² + 6(-3) + 5
y = 9 – 18 + 5 = -4
Thus, the vertex is at the point (-3, -4).
2. Axis of Symmetry
The axis of symmetry is a vertical line that passes through the vertex. It can be written as:
x = -3
This means the graph is symmetric around this line.
3. Intercepts
To find the intercepts, we will calculate both the x-intercepts and y-intercept.
A. Y-Intercept
The y-intercept occurs where x = 0. Substitute x = 0 into the equation:
y = (0)² + 6(0) + 5 = 5
Thus, the y-intercept is at the point (0, 5).
B. X-Intercepts
The x-intercepts occur where y = 0. To find these, set the equation to zero:
0 = x² + 6x + 5
This can be factored as:
0 = (x + 1)(x + 5)
Setting each factor to zero gives:
x + 1 = 0 → x = -1
x + 5 = 0 → x = -5
Thus, the x-intercepts are at the points (-1, 0) and (-5, 0).
Summary
In conclusion:
- Vertex: (-3, -4)
- Axis of Symmetry: x = -3
- Y-Intercept: (0, 5)
- X-Intercepts: (-1, 0) and (-5, 0)