Graffiau cwadratig – Haen Ganolradd ac Uwch
\(\text{y = 5x}^2~\text{+~x~+~1}\)
Graff cwadratig yw unrhyw graff sydd â \(\text{x}^2\) yn ei hafaliad. Rydyn ni’n eu llunio mewn ffordd debyg iawn i graffiau llinell syth a bydd angen i ni amnewid gwerthoedd yn yr hafaliad. Bydd pob graff cwadratig ar ffurf cromlin.
Enghraifft
Rydyn ni eisiau llunio graff \(\text{y = x}^2~{+~3}\) felly bydd angen i ni gwblhau’r tabl gwerthoedd hwn:
\(\text{x}\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
\(\text{y = x}^2~{+~3}\) |
\(\text{x}\) |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
\(\text{y = x}^2~{+~3}\) |
- pan fo \(\text{x}\) = -3, \(\text{y}\) = (-3 × -3) + 3 = 12
- pan fo \(\text{x}\) = -2, \(\text{y}\) = (-2 × -2) + 3 = 7
- pan fo \(\text{x}\) = -1, \(\text{y}\) = (-1 × -1) + 3 = 4
- pan fo \(\text{x}\) = 0, \(\text{y}\) = (0 × 0) + 3 = 3
- pan fo \(\text{x}\) = 1, \(\text{y}\) = (1 × 1) + 3 = 4
- pan fo \(\text{x}\) = 2, \(\text{y}\) = (2 × 2) + 3 = 7
- pan fo \(\text{x}\) = 3, \(\text{y}\) = (3 × 3) + 3 = 12
Felly bydd ein tabl gwerthoedd gorffenedig yn edrych fel hyn:
\(\text{x}\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
\(\text{y = x}^2~{+~3}\) | 12 | 7 | 4 | 3 | 4 | 7 | 12 |
\(\text{x}\) |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
\(\text{y = x}^2~{+~3}\) |
12 |
7 |
4 |
3 |
4 |
7 |
12 |
I lunio’r graff hwn, rhaid i ni feddwl am y gwerthoedd yn y tabl hwn fel cyfesurynnau.
Felly cyfesurynnau’r pwynt cyntaf fydd \(\text{(-3,~12)}\), cyfesurynnau’r ail fydd \(\text{(-2,~7)}\) ayyb.
Yna rydyn ni’n uno’r pwyntiau â chromlin:
Question
Cwblha’r tabl a llunia graff \(\text{y = 2x}^2~{-~1}\).
\(\text{x}\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
\(\text{y = 2x}^2~{-~1}\) |
\(\text{x}\) |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
\(\text{y = 2x}^2~{-~1}\) |
- pan fo \(\text{x}\) = -3, \(\text{y}\) = (2 × -3 × -3) - 1 = 17
- pan fo \(\text{x}\) = -2, \(\text{y}\) = (2 × -2 × -2) - 1 = 7
- pan fo \(\text{x}\) = -1, \(\text{y}\) = (2 × -1 × -1) - 1 = 1
- pan fo \(\text{x}\) = 0, \(\text{y}\) = (2 × 0 × 0) - 1 = -1
- pan fo \(\text{x}\) = 1, \(\text{y}\) = (2 × 1 × 1) - 1 = 1
- pan fo \(\text{x}\) = 2, \(\text{y}\) = (2 × 2 × 2) - 1 = 7
- pan fo \(\text{x}\) = 3, \(\text{y}\) = (2 × 3 × 3) - 1 = 17
\(\text{x}\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
\(\text{y = 2x}^2~{-~1}\) | 17 | 7 | 1 | -1 | 1 | 7 | 17 |
\(\text{x}\) |
-3 |
-2 |
-1 |
0 |
1 |
2 |
3 |
\(\text{y = 2x}^2~{-~1}\) |
17 |
7 |
1 |
-1 |
1 |
7 |
17 |