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Sagot :
substituteGiven:-
[tex](0,0),(-3,0)(-1,3)[/tex]To find the quadratic equation.
So now we use the formula,
[tex]y=ax^2+bx+c[/tex]So now we subtitute the points and find the value of a,b,c. So we get,
[tex]\begin{gathered} 0=a(0)+b(0)+c \\ c=0 \end{gathered}[/tex]Also,
[tex]\begin{gathered} 0=a(-3)^2+b(-3)+c \\ 0=9a-3b \end{gathered}[/tex]Also,
[tex]\begin{gathered} 3=a(-1)^2+b(-1)+0 \\ 3=a-b \end{gathered}[/tex]So now we simplify both equation. so we get,
[tex]\begin{gathered} 9a-3b=0 \\ 3a-3b=9 \end{gathered}[/tex]Now we add both the equations. we get,
[tex]\begin{gathered} 6a=-9 \\ a=-\frac{3}{2} \end{gathered}[/tex]Now we find the value of b, so we get,
[tex]\begin{gathered} a-b=3 \\ -\frac{3}{2}-b=3 \\ -b=3+\frac{3}{2} \\ -b=\frac{9}{2} \\ b=-\frac{9}{2} \end{gathered}[/tex]So the required values are,
[tex]y=-\frac{3}{2}x^2-\frac{9}{2}x+0[/tex]
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