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Sagot :
The general form of a quadratic is:
[tex]y=ax^2+bx+c[/tex]We need to plug in the 3 pair of points into "x" and "y" and simultaneously solve the 3 equations for a, b, and c.
Putting (0,9):
[tex]\begin{gathered} 9=a(0)^2+b(0)+c \\ c=9 \end{gathered}[/tex]Putting (-6,9):
[tex]\begin{gathered} 9=a(-6)^2+b(-6)+c \\ 9=36a-6b+9 \\ 36a=6b \\ a=\frac{6b}{36} \\ a=\frac{b}{6} \end{gathered}[/tex]Putting (-5,4):
[tex]\begin{gathered} 4=a(-5)^2+b(-5)+9 \\ 4=25a-5b+9 \\ 25a-5b=-5 \end{gathered}[/tex]We substitute a=b/6 into this equation and solve for b:
[tex]\begin{gathered} 25(\frac{b}{6})-5b=-5 \\ \frac{25b}{6}-5b=-5 \\ \frac{25b-30b}{6}=-5 \\ -5b=-30 \\ b=6 \end{gathered}[/tex]Thus, "a" will be:
[tex]\begin{gathered} a=\frac{b}{6} \\ a=\frac{6}{6} \\ a=1 \end{gathered}[/tex]Thus, we have a = 1, b = 6, c = 9
The equation is:
[tex]\begin{gathered} y=1x^2+6x+9 \\ y=x^2+6x+9 \end{gathered}[/tex]
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