Answer :
To draw the graph of the quadratic function [tex]\( f(x) = x^2 + x - 2 \)[/tex], we need to understand its key features such as its vertex, axis of symmetry, roots (x-intercepts), y-intercept, and general shape. Here's a detailed, step-by-step solution:
### 1. Determine the Roots (x-intercepts)
To find the roots of the function [tex]\( f(x) = x^2 + x - 2 \)[/tex], we set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^2 + x - 2 = 0 \][/tex]
To solve this quadratic equation, we can factorize it:
[tex]\[ (x+2)(x-1) = 0 \][/tex]
Setting each factor to zero gives us the roots:
[tex]\[ x+2 = 0 \Rightarrow x = -2 \][/tex]
[tex]\[ x-1 = 0 \Rightarrow x = 1 \][/tex]
So, the function intersects the x-axis at [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex].
### 2. Determine the Vertex
The vertex of a parabola given by [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For our function [tex]\( f(x) = x^2 + x - 2 \)[/tex], [tex]\( a = 1 \)[/tex] and [tex]\( b = 1 \)[/tex]:
[tex]\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = -\frac{1}{2} \)[/tex] back into the function:
[tex]\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 2 \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 2 \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1 - 2 - 8}{4} \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{-9}{4} \][/tex]
So, the vertex of the parabola is [tex]\(\left(-\frac{1}{2}, -\frac{9}{4}\right)\)[/tex].
### 3. Determine the y-intercept
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] and solving for [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 0^2 + 0 - 2 = -2 \][/tex]
So, the function intersects the y-axis at [tex]\( (0, -2) \)[/tex].
### 4. Sketch the Parabola
Now, we have enough information to sketch the parabola:
1. The roots (x-intercepts) are at [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
2. The vertex is at [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex].
3. The y-intercept is at [tex]\( (0, -2) \)[/tex].
### 5. Plotting the Key Points and the Curve
1. Plot the x-intercepts [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
2. Plot the vertex [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex].
3. Plot the y-intercept [tex]\( (0, -2) \)[/tex].
Finally, draw a smooth curve through these points that opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive), ensuring the parabola is symmetrical around the vertical line passing through the vertex (x = -1/2).
### Summary
Here's the graph of the quadratic function [tex]\( f(x) = x^2 + x - 2 \)[/tex]:
```
y
^
|
|
|
|
|
| *
|
____________________|_________________________________> x
-2 1
```
The key points are:
- Vertex: [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex]
- X-intercepts: [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
- Y-intercept: [tex]\( (0, -2) \)[/tex]
This gives you a visual representation and the key characteristics of the quadratic function [tex]\( f(x) = x^2 + x - 2 \)[/tex].
### 1. Determine the Roots (x-intercepts)
To find the roots of the function [tex]\( f(x) = x^2 + x - 2 \)[/tex], we set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ x^2 + x - 2 = 0 \][/tex]
To solve this quadratic equation, we can factorize it:
[tex]\[ (x+2)(x-1) = 0 \][/tex]
Setting each factor to zero gives us the roots:
[tex]\[ x+2 = 0 \Rightarrow x = -2 \][/tex]
[tex]\[ x-1 = 0 \Rightarrow x = 1 \][/tex]
So, the function intersects the x-axis at [tex]\( x = -2 \)[/tex] and [tex]\( x = 1 \)[/tex].
### 2. Determine the Vertex
The vertex of a parabola given by [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For our function [tex]\( f(x) = x^2 + x - 2 \)[/tex], [tex]\( a = 1 \)[/tex] and [tex]\( b = 1 \)[/tex]:
[tex]\[ x = -\frac{1}{2 \cdot 1} = -\frac{1}{2} \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = -\frac{1}{2} \)[/tex] back into the function:
[tex]\[ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) - 2 \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 2 \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} - \frac{8}{4} \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{1 - 2 - 8}{4} \][/tex]
[tex]\[ f\left(-\frac{1}{2}\right) = \frac{-9}{4} \][/tex]
So, the vertex of the parabola is [tex]\(\left(-\frac{1}{2}, -\frac{9}{4}\right)\)[/tex].
### 3. Determine the y-intercept
The y-intercept is found by setting [tex]\( x = 0 \)[/tex] and solving for [tex]\( f(x) \)[/tex]:
[tex]\[ f(0) = 0^2 + 0 - 2 = -2 \][/tex]
So, the function intersects the y-axis at [tex]\( (0, -2) \)[/tex].
### 4. Sketch the Parabola
Now, we have enough information to sketch the parabola:
1. The roots (x-intercepts) are at [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
2. The vertex is at [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex].
3. The y-intercept is at [tex]\( (0, -2) \)[/tex].
### 5. Plotting the Key Points and the Curve
1. Plot the x-intercepts [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
2. Plot the vertex [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex].
3. Plot the y-intercept [tex]\( (0, -2) \)[/tex].
Finally, draw a smooth curve through these points that opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive), ensuring the parabola is symmetrical around the vertical line passing through the vertex (x = -1/2).
### Summary
Here's the graph of the quadratic function [tex]\( f(x) = x^2 + x - 2 \)[/tex]:
```
y
^
|
|
|
|
|
| *
|
____________________|_________________________________> x
-2 1
```
The key points are:
- Vertex: [tex]\( \left(-\frac{1}{2}, -\frac{9}{4}\right) \)[/tex]
- X-intercepts: [tex]\( (-2, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex]
- Y-intercept: [tex]\( (0, -2) \)[/tex]
This gives you a visual representation and the key characteristics of the quadratic function [tex]\( f(x) = x^2 + x - 2 \)[/tex].
