Complete a table: quadratic equations
Key notes !
1. Understanding Quadratic Equations
A quadratic equation is a second-degree polynomial equation of the form:
f(x) = ax^2 + bx + c
Where:
- ( a ), ( b ), and ( c ) are constants.
- ( a not eqauls to 0 ).
2. Purpose of Creating a Table
Creating a table helps to:
- Visualize the behavior of the quadratic function.
- Identify key points such as the vertex, axis of symmetry, roots (if any), and the y-intercept.
- Understand how the function changes as ( x ) varies.
3. Steps to Create a Table for a Quadratic Equation
Step 1: Choose a Range of ( x ) Values
Select a range of ( x ) values that are evenly spaced and cover both negative and positive values. This ensures that the table captures the behavior of the quadratic function around its vertex and across the x-axis.
For example, if the quadratic equation is ( f(x) = 2x^2 – 3x + 1 ), you might choose ( x ) values such as (-2, -1, 0, 1, 2).
Step 2: Substitute ( x ) Values into the Equation
For each chosen ( x ) value, substitute it into the quadratic equation to find the corresponding ( f(x) ) value. This involves simple arithmetic to evaluate the expression.
Step 3: Calculate ( f(x) ) for Each ( x ) Value
Perform the calculation for each ( x ) value and record the result.
4. Example of Completing a Table
Let’s consider the quadratic equation:
f(x) = x^2 – 4x + 3
Step-by-Step Table Creation
- Select ( x ) values: Choose values around the expected vertex, such as ( -1, 0, 1, 2, 3, 4, 5 ).
- Create the table:
( x ) | Calculation | ( f(x) ) |
---|---|---|
-1 | ((-1)^2 – 4(-1) + 3) | ( 1 + 4 + 3 = 8 ) |
0 | (0^2 – 4(0) + 3) | ( 0 + 0 + 3 = 3 ) |
1 | (1^2 – 4(1) + 3) | ( 1 – 4 + 3 = 0 ) |
2 | (2^2 – 4(2) + 3) | ( 4 – 8 + 3 = -1 ) |
3 | (3^2 – 4(3) + 3) | ( 9 – 12 + 3 = 0 ) |
4 | (4^2 – 4(4) + 3) | ( 16 – 16 + 3 = 3 ) |
5 | (5^2 – 4(5) + 3) | ( 25 – 20 + 3 = 8 ) |
Final Table:
( x ) | ( f(x) ) |
---|---|
-1 | 8 |
0 | 3 |
1 | 0 |
2 | -1 |
3 | 0 |
4 | 3 |
5 | 8 |
5. Interpreting the Table
- Vertex: The lowest (or highest for negative (a)) point on the table is the vertex. In this case, at ( x = 2 ), ( f(x) = -1 ) suggests the vertex is at (2, -1).
- Axis of Symmetry: The x-value of the vertex (x = 2) acts as the line of symmetry for the parabola.
- Roots: The x-values where ( f(x) = 0 ) are the roots of the equation. In this case, ( x = 1 ) and ( x = 3 ).
- Y-Intercept: The value of ( f(x) ) when ( x = 0 ) is the y-intercept. Here, it’s 3.
6. Graphing Using the Table
Plotting these ( (x, f(x)) ) points on a coordinate plane helps visualize the parabolic shape of the quadratic equation. You can use the table data to sketch the curve, marking key points like the vertex, intercepts, and roots.
Learn with an example
Complete the table.
f(p) = p2 − 4 | |
p | f(p) |
-2 | |
-1 | |
0 | |
1 |
The first p-value in the table is -2. Evaluate f(p) = p2 − 4 for p = -2.
f (p) = p2 – 4.
= (-2)2 – 4 plug in p = -2
= 4 – 4 Square
= 0 Subtract
When p = -2, f(p) = 0. Complete the first row of the table.
f(p) = p2 − 4 | |
p | f(p) |
-2 | 0 |
-1 | ? |
0 | ? |
1 | ? |
Next, evaluate f(p) = p2 − 4 for p = -1.
f (p) = p2 – 4.
= (-1)2 – 4 plug in p = -1
= 1 – 4 Square
= -3 Subtract
When p = -1, f(p) = -3. Complete the second row of the table.
f(p) = p2 − 4 | |
p | f(p) |
-2 | 0 |
-1 | -3 |
0 | ? |
1 | ? |
Complete the rest of the table the same way.
f(p) = p2 − 4 | |
p | f(p) |
-2 | 0 |
-1 | -3 |
0 | -4 |
1 | -3 |
Complete the table.
f(r) = r2 + 10 | |
r | f(r) |
-1 | |
0 | |
1 | |
2 |
The first R-value in the table is -1. Evaluate f(r) = r2 + 10 for r = -1.
f (r) = r2 +10
= (-1)2 + 10 plug in r = -1
= 1 + 10 Square
= 11 Add
When r = -1, f(r) = 11. Complete the first row of the table.
f(r) = r2 + 10 | |
r | f(r) |
-1 | 11 |
0 | ? |
1 | ? |
2 | ? |
Next, evaluate f(r) = r2 + 10 for r = 0.
f (r) = r2 +10
= (0)2 + 10 plug in r = 0
= 0 + 10 Square
= 10 Add
When r = 0, f(r) = 10. Complete the second row of the table.
f(r) = r2 + 10 | |
r | f(r) |
-1 | 11 |
0 | 10 |
1 | ? |
2 | ? |
Complete the rest of the table the same way.
f(r) = r2 + 10 | |
r | f(r) |
-1 | 11 |
0 | 10 |
1 | 11 |
2 | 14 |
Complete the table.
f(j) = j2 | |
j | f(j) |
-2 | |
-1 | |
0 | |
1 |
The first j-value in the table is -2. Evaluate f(j) = j2 for j = -2.
f (j) = j2
= (-2 )2 plug in j = – 2
= 4 Square
When j = -2, f(j) = 4. Complete the first row of the table.
f(j) = j2 | |
j | f(j) |
-2 | 4 |
-1 | ? |
0 | ? |
1 | ? |
Next, evaluate f(j) = j2 for j = -1.
f (j) = j2
= (-1 )2 plug in j = – 1
= 1 Square
When j = -1, f(j) = 1. Complete the second row of the table.
f(j) = j2 | |
j | f(j) |
-2 | 4 |
-1 | 1 |
0 | ? |
1 | 0 |
Complete the rest of the table the same way.
f(j) = j2 | |
j | f(j) |
-2 | 4 |
-1 | 1 |
0 | 0 |
1 | 1 |
let’s practice!