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Graphing a Quadratic Function Given in Factored Form
Plot the function
y=(x+1)(x-3) by following these steps:
Step 1: Plot the x-intercepts of the function.
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Intro
ntViewers/FrameChain/Activity#_
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Sagot :

Absolutely, let's go through the process of plotting the function [tex]\( y = (x + 1)(x - 3) \)[/tex] step by step.

### Step 1: Find the x-intercepts
The x-intercepts are the points where the graph of the function crosses the x-axis. This occurs when [tex]\( y = 0 \)[/tex].

Given the equation:
[tex]\[ y = (x + 1)(x - 3) \][/tex]

Set [tex]\( y \)[/tex] to 0:
[tex]\[ (x + 1)(x - 3) = 0 \][/tex]

Solve for [tex]\( x \)[/tex]:
[tex]\[ x + 1 = 0 \quad \text{or} \quad x - 3 = 0 \][/tex]
[tex]\[ x = -1 \quad \text{or} \quad x = 3 \][/tex]

Thus, the x-intercepts are:
[tex]\[ x = -1 \text{ and } x = 3 \][/tex]

### Step 2: Plot the x-intercepts on the graph
The x-intercepts are the points where the graph crosses the x-axis:
[tex]\[ (-1, 0) \text{ and } (3, 0) \][/tex]

### Step 3: Find the vertex
The vertex of a parabola in factored form can be found as follows:
1. The x-coordinate of the vertex is the midpoint of the x-intercepts:
[tex]\[ x_{\text{vertex}} = \frac{x_1 + x_2}{2} \][/tex]
Given the x-intercepts [tex]\( x_1 = -1 \)[/tex] and [tex]\( x_2 = 3 \)[/tex]:
[tex]\[ x_{\text{vertex}} = \frac{-1 + 3}{2} = 1 \][/tex]

2. Substitute [tex]\( x_{\text{vertex}} \)[/tex] back into the equation to find the y-coordinate:
[tex]\[ y_{\text{vertex}} = (1 + 1)(1 - 3) \][/tex]
[tex]\[ y_{\text{vertex}} = 2 \cdot (-2) = -4 \][/tex]

Hence, the vertex is:
[tex]\[ (1, -4) \][/tex]

### Plotting on the graph:

1. Plot the x-intercepts [tex]\( (-1, 0) \)[/tex] and [tex]\( (3, 0) \)[/tex].
2. Plot the vertex [tex]\( (1, -4) \)[/tex].

### Sketch the function
Using these points, we can now sketch the quadratic function:

- Start at one x-intercept (e.g., [tex]\( (-1, 0) \)[/tex]).
- Draw the curve downward to the vertex [tex]\( (1, -4) \)[/tex].
- Continue the curve upward to the other x-intercept [tex]\( (3, 0) \)[/tex].

This will give you a parabola that opens downward (since the coefficient of the [tex]\( x^2 \)[/tex] term is negative when expanded).

### Final Graph
Your graph should look like a 'U' shaped curve that is flipped upside down, crossing the x-axis at [tex]\((-1, 0)\)[/tex] and [tex]\( (3, 0) \)[/tex] and reaching its lowest point (vertex) at [tex]\( (1, -4) \)[/tex].

Good luck on your graphing! Make sure to label the intercepts and the vertex for clarity.