Quadratic form is given in the form
[tex]y= ax^{2}+bx+c [/tex], where a, b, and c are constants
We substitute the value (8, 12) and (5,3) in turn
[tex]12= a(8)^{2}+b(8)+c[/tex]
[tex]12=64a+8b+c[/tex] ... Equation 1
[tex]3=a(5)^{2}+b(5)+c [/tex]
[tex]3=25a+5b+c[/tex] ... Equation 2
Equation 1 - Equation 2 gives
[tex]39a+3b=9[/tex], we call this Equation 3
We also know that (5, 3) is the turning point of the curve, where the value of [tex]x[/tex] is given by [tex]x=- \frac{b}{2a} [/tex]
Substitute [tex]x=5[/tex]
[tex]5=- \frac{b}{2a} [/tex]
[tex]10a=-b[/tex]
[tex]b=-10a[/tex] ... Equation 4
Substitute Equation 4 into Equation 3
[tex]39a+3b=9[/tex]
[tex]39a+3(-10a)=9[/tex]
[tex]39a-30a=9[/tex]
[tex]9a=9[/tex]
[tex]a=1[/tex]
Now we need to work out the value of [tex]b[/tex] and [tex]c[/tex]
Substitute [tex]a=1[/tex] back into equation 4
[tex]-10a=b[/tex]
[tex]-10(1)=b[/tex]
[tex]b=-10[/tex]
Substitute [tex]a=1[/tex] and [tex]b=-10[/tex] into either equation 1 or equation 2
[tex]64a+8b+c=12[/tex]
[tex]64(1)+8(-10)+c=12[/tex]
[tex]64-80+c=12[/tex]
[tex]c=12+80-64[/tex]
[tex]c=28[/tex]
Hence the equation of the quadratic curve is
[tex]y= x^{2} 10x+28[/tex]
