Answer :
Let's go through the steps to graph the quadratic function [tex]\( f(x) = (x + 1)(x - 5) \)[/tex]:
1. Identify the [tex]\(x\)[/tex]-intercepts:
To find the [tex]\(x\)[/tex]-intercepts, we set [tex]\(f(x) = 0\)[/tex]. This gives us:
[tex]\[ (x + 1)(x - 5) = 0 \][/tex]
Solving this equation, we get:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x - 5 = 0 \][/tex]
Thus, [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
Hence, the [tex]\(x\)[/tex]-intercepts are [tex]\((-1, 0)\)[/tex] and [tex]\((5, 0)\)[/tex].
2. Find the midpoint between the intercepts:
The [tex]\(x\)[/tex]-coordinate of the midpoint between [tex]\((-1, 0)\)[/tex] and [tex]\((5, 0)\)[/tex] is found by taking the average of the [tex]\(x\)[/tex]-coordinates:
[tex]\[ \text{Midpoint}_x = \frac{-1 + 5}{2} = \frac{4}{2} = 2 \][/tex]
Therefore, the midpoint on the x-axis is [tex]\((2, 0)\)[/tex].
3. Find the vertex:
For the quadratic function in factored form, the vertex [tex]\(x\)[/tex]-coordinate is the same as the midpoint's [tex]\(x\)[/tex]-coordinate. So the [tex]\(x\)[/tex]-coordinate of the vertex is [tex]\(2\)[/tex].
To find the [tex]\(y\)[/tex]-coordinate of the vertex, substitute [tex]\(x = 2\)[/tex] back into the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(2) = (2 + 1)(2 - 5) = 3 \times (-3) = -9 \][/tex]
Therefore, the vertex of the function is [tex]\((2, -9)\)[/tex].
4. Find the [tex]\(y\)[/tex]-intercept:
The [tex]\(y\)[/tex]-intercept is found by setting [tex]\(x = 0\)[/tex] in the function:
[tex]\[ f(0) = (0 + 1)(0 - 5) = 1 \times (-5) = -5 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is [tex]\((0, -5)\)[/tex].
5. Plot another point, then draw the graph:
To ensure the accuracy of the graph, we can calculate another point. Let's use [tex]\(x = 1\)[/tex]:
[tex]\[ f(1) = (1 + 1)(1 - 5) = 2 \times (-4) = -8 \][/tex]
So, another point on the graph is [tex]\((1, -8)\)[/tex].
Based on these calculations, our results are:
- [tex]\(x\)[/tex]-intercepts: [tex]\((-1, 0)\)[/tex], [tex]\((5, 0)\)[/tex]
- Midpoint: [tex]\((2, 0)\)[/tex]
- Vertex: [tex]\((2, -9)\)[/tex]
- [tex]\(y\)[/tex]-intercept: [tex]\((0, -5)\)[/tex]
- Another point: [tex]\((1, -8)\)[/tex]
Using these points, you can now plot the quadratic function [tex]\(f(x) = (x + 1)(x - 5)\)[/tex] accurately on a graph.
1. Identify the [tex]\(x\)[/tex]-intercepts:
To find the [tex]\(x\)[/tex]-intercepts, we set [tex]\(f(x) = 0\)[/tex]. This gives us:
[tex]\[ (x + 1)(x - 5) = 0 \][/tex]
Solving this equation, we get:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x - 5 = 0 \][/tex]
Thus, [tex]\( x = -1 \)[/tex] and [tex]\( x = 5 \)[/tex].
Hence, the [tex]\(x\)[/tex]-intercepts are [tex]\((-1, 0)\)[/tex] and [tex]\((5, 0)\)[/tex].
2. Find the midpoint between the intercepts:
The [tex]\(x\)[/tex]-coordinate of the midpoint between [tex]\((-1, 0)\)[/tex] and [tex]\((5, 0)\)[/tex] is found by taking the average of the [tex]\(x\)[/tex]-coordinates:
[tex]\[ \text{Midpoint}_x = \frac{-1 + 5}{2} = \frac{4}{2} = 2 \][/tex]
Therefore, the midpoint on the x-axis is [tex]\((2, 0)\)[/tex].
3. Find the vertex:
For the quadratic function in factored form, the vertex [tex]\(x\)[/tex]-coordinate is the same as the midpoint's [tex]\(x\)[/tex]-coordinate. So the [tex]\(x\)[/tex]-coordinate of the vertex is [tex]\(2\)[/tex].
To find the [tex]\(y\)[/tex]-coordinate of the vertex, substitute [tex]\(x = 2\)[/tex] back into the function [tex]\(f(x)\)[/tex]:
[tex]\[ f(2) = (2 + 1)(2 - 5) = 3 \times (-3) = -9 \][/tex]
Therefore, the vertex of the function is [tex]\((2, -9)\)[/tex].
4. Find the [tex]\(y\)[/tex]-intercept:
The [tex]\(y\)[/tex]-intercept is found by setting [tex]\(x = 0\)[/tex] in the function:
[tex]\[ f(0) = (0 + 1)(0 - 5) = 1 \times (-5) = -5 \][/tex]
So, the [tex]\(y\)[/tex]-intercept is [tex]\((0, -5)\)[/tex].
5. Plot another point, then draw the graph:
To ensure the accuracy of the graph, we can calculate another point. Let's use [tex]\(x = 1\)[/tex]:
[tex]\[ f(1) = (1 + 1)(1 - 5) = 2 \times (-4) = -8 \][/tex]
So, another point on the graph is [tex]\((1, -8)\)[/tex].
Based on these calculations, our results are:
- [tex]\(x\)[/tex]-intercepts: [tex]\((-1, 0)\)[/tex], [tex]\((5, 0)\)[/tex]
- Midpoint: [tex]\((2, 0)\)[/tex]
- Vertex: [tex]\((2, -9)\)[/tex]
- [tex]\(y\)[/tex]-intercept: [tex]\((0, -5)\)[/tex]
- Another point: [tex]\((1, -8)\)[/tex]
Using these points, you can now plot the quadratic function [tex]\(f(x) = (x + 1)(x - 5)\)[/tex] accurately on a graph.
