Let's go through the steps to graph the function [tex]\( f(x) = (x + 1)(x - 5) \)[/tex]. We will break it down step-by-step.
### Step 1: Identify the [tex]\( x \)[/tex]-intercepts
The [tex]\( x \)[/tex]-intercepts can be found by setting [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ (x + 1)(x - 5) = 0 \][/tex]
This gives us two roots:
[tex]\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \][/tex]
[tex]\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \][/tex]
Thus, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ (-1, 0) \text{ and } (5, 0) \][/tex]
### Step 2: Find the midpoint between the intercepts
The midpoint [tex]\( x \)[/tex]-coordinate between [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex] can be found using the midpoint formula:
[tex]\[ \text{Midpoint}_x = \frac{-1 + 5}{2} = 2.0 \][/tex]
So the [tex]\( x \)[/tex]-coordinate of the midpoint is [tex]\( 2.0 \)[/tex].
### Step 3: Find the vertex
The vertex of the quadratic function in factored form can be found by evaluating the function at the midpoint:
[tex]\[ f(2.0) = (2 + 1)(2 - 5) = 3 \cdot (-3) = -9 \][/tex]
Thus, the vertex is:
[tex]\[ (2.0, -9.0) \][/tex]
### Step 4: Find the [tex]\( y \)[/tex]-intercept
The [tex]\( y \)[/tex]-intercept can be found by evaluating the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = (0 + 1)(0 - 5) = 1 \cdot (-5) = -5 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept is:
[tex]\[ (0, -5) \][/tex]
### Step 5: Plot another point
We can choose another point to ensure the accuracy of the graph. Let's use [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = (2 + 1)(2 - 5) = 3 \cdot (-3) = -9 \][/tex]
So,
[tex]\[ (2, -9) \][/tex]
### Conclusion
When graphing the function [tex]\( f(x) = (x + 1)(x - 5) \)[/tex], use the following points:
- [tex]\( x \)[/tex]-intercepts: [tex]\( (-1, 0) \)[/tex] and [tex]\( (5, 0) \)[/tex]
- Vertex: [tex]\( (2.0, -9.0) \)[/tex]
- [tex]\( y \)[/tex]-intercept: [tex]\( (0, -5) \)[/tex]
- Another point: [tex]\( (2, -9) \)[/tex]
Using these points, you can sketch the parabola on the coordinate plane. The vertex will be at [tex]\( (2.0, -9.0) \)[/tex], ensuring that the graph opens upwards since the leading coefficient (from expansion would be [tex]\( x^2 + bx + c \)[/tex]) is positive. The graph should pass through the identified intercepts and points.
