To determine which graph represents the equation [tex]\((x+1)(x+5)=0\)[/tex], let's first understand the nature of the equation. This equation is a quadratic equation and can be written in its expanded form:
[tex]\[ (x+1)(x+5) = 0 \][/tex]
To solve for [tex]\(x\)[/tex], we set each factor equal to zero:
1. [tex]\(x+1 = 0\)[/tex]
2. [tex]\(x+5 = 0\)[/tex]
Solving these, we get:
1. [tex]\(x = -1\)[/tex]
2. [tex]\(x = -5\)[/tex]
Therefore, the solutions to the equation [tex]\((x+1)(x+5)=0\)[/tex] are [tex]\(x = -1\)[/tex] and [tex]\(x = -5\)[/tex]. These values are the x-intercepts of the graph of the equation.
Next, let's consider the shape of the graph. Since the equation is a quadratic and it can be expanded to the standard form [tex]\(Ax^2 + Bx + C = 0\)[/tex] (specifically, this expands to [tex]\(x^2 + 6x + 5 = 0\)[/tex]), the graph will be a parabola. The coefficient of [tex]\(x^2\)[/tex] is positive (1), so the parabola opens upwards.
Based on this information, the key characteristics of the graph should be:
1. It is a parabola opening upwards.
2. It has x-intercepts at [tex]\(x = -1\)[/tex] and [tex]\(x = -5\)[/tex].
To identify the correct graph, look for a parabola that intersects the x-axis at [tex]\(x = -1\)[/tex] and [tex]\(x = -5\)[/tex]. The vertex of this parabola will be midway between these intercepts, giving it a vertex along the x-axis at [tex]\(x = -3\)[/tex], and since it opens upwards, it will have its minimum point at the vertex.
So, the graph you are looking for is a parabola that:
- Opens upwards.
- Crosses the x-axis at [tex]\(x = -1\)[/tex] and [tex]\(x = -5\)[/tex].
This graph correctly represents the equation [tex]\((x+1)(x+5)=0\)[/tex].
