Complete a table: quadratic equations

Key notes !

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 ).

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.

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).

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.

Perform the calculation for each ( x ) value and record the result.

Let’s consider the quadratic equation:


f(x) = x^2 – 4x + 3

  1. Select ( x ) values: Choose values around the expected vertex, such as ( -1, 0, 1, 2, 3, 4, 5 ).
  2. 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) )
-18
03
10
2-1
30
43
58
  • 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.

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

f(p) = p2 − 4
pf(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
pf(p)
-20
-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
pf(p)
-20
-1-3
0?
1?

Complete the rest of the table the same way.

f(p) = p2 − 4
pf(p)
-20
-1-3
0-4
1-3
f(r) = r2 + 10
rf(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
rf(r)
-111
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
rf(r)
-111
010
1?
2?

Complete the rest of the table the same way.

f(r) = r2 + 10
rf(r)
-111
010
111
214

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
jf(j)
-24
-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
jf(j)
-24
-11
0?
10

Complete the rest of the table the same way.

f(j) = j2
jf(j)
-24
-11
00
11

let’s practice!