[tex]\displaystyle\\Answer:\ y=\frac{7}{16}x^2 -\frac{7}{4}x-\frac{5}{4}[/tex]
Step-by-step explanation:
The vertex is also the symmetry point of the parabola. The formula for finding the x-coordinate of the parabola: x = -b/2a (2,-3)
Hence,
[tex]\displaystyle\\2=\frac{-b}{2a} \\[/tex]
Multiply both parts of the equation by -2a:
[tex]\displaystyle\\-4a=b\ \ \ \ \ (1)[/tex]
[tex]y=ax^2+bx+c\ \ \ \ \ -\ \ \ \ \ the\ quadratic\ equation\\\\Thus,[/tex]
You can make a system of equations on two points belonging to the quadratic equation:
[tex]-3=a(2)^2+b(2)+c\\4=a(6)^2+b(6)+c\\\\-3=4a+2b+c\ \ \ \ (2)\\4=36a+6b+c\ \ \ \ (3)\\\\[/tex]
Substitute (1) into equations (2) and (3):
[tex]-3=4a+2(-4a)+c\\4=36a+6(-4a)+c\\\\-3=4a-8a+c\\4=36a-24a+c\\\\-3=-4a+c\ \ \ \ (4)\\4=12a+c \ \ \ \ (5)\\\\\\[/tex]
Subtract equation (4) from equation (5):
[tex]7=16a[/tex]
Divide both parts of the equation by 16:
[tex]\displaystyle\\\frac{7}{16} =a\ \ \ \ (6)[/tex]
Substitute (6) into equations (1):
[tex]\displaystyle\\-4(\frac{7}{16} )=b\\\\-\frac{4*7}{4*4}=b\\\\-\frac{7}{4}=b[/tex]
Substitute values a and b into equation (2):
[tex]\displaystyle-3=4(\frac{7}{16})+2(-\frac{7}{4})+c\\\\ -3=\frac{7}{4} -\frac{14}{4}+c\\\\ -3=-\frac{7}{4}+c \\\\-3+\frac{7}{4}=-\frac{7}{4}+c+\frac{7}{4} \\\\ \frac{-3*4+7}{4} =c\\\\\frac{-12+7}{4}=c\\\\ -\frac{5}{4}=c[/tex]
Thus,
[tex]\displaystyle\\y=\frac{7}{16}x^2 -\frac{7}{4}x-\frac{5}{4}[/tex]
