we are given the following equation:
[tex]y=ax^2+bx+c[/tex]We are also given the points:
[tex]\begin{gathered} (-2,-13) \\ (2,3) \\ (4,5) \end{gathered}[/tex]replacing the points:
[tex]\begin{gathered} -13=a(-2)^2+b(-2)+c \\ 3=a(2)^2+b(2)+c \\ 5=a(4)^2+b(4)+c \end{gathered}[/tex]Simplifying:
[tex]\begin{gathered} 4a-2b+c=-13,(1) \\ 4a+2b+c=3,(2) \\ 16a+4b+c=5,(3) \end{gathered}[/tex]We get three equations and three variables. To solve the system we will add equations (1) and (2):
[tex]\begin{gathered} 4a-2b+c+4a+2b+c=-13+3 \\ 8a+2c=-10,(4) \end{gathered}[/tex]Now we multiply equation (2)by -2 and add that to equation (3):
[tex]\begin{gathered} -2(4a+2b+c)+16a+4b+c=-2(3)+5 \\ -8a-4b-2c+16a+4b+c=-6+5 \\ 8a-c=-1,(5) \end{gathered}[/tex]Now we multiply equation(4) by -1 and add that to equation (5):
[tex]\begin{gathered} -(8a+2c)+8a-c=-(-10)-1 \\ -8a-2c+8a-c=10-1 \\ -3c=9 \\ c=\frac{9}{-3}=-3 \end{gathered}[/tex]Therefore, c = -3. replacing the value of "c" in equation (5):
[tex]\begin{gathered} 8a-(-3)=-1 \\ 8a+3=-1 \end{gathered}[/tex]subtracting 3 to both sides:
[tex]\begin{gathered} 8a=-1-3 \\ 8a=-4 \end{gathered}[/tex]Dividing both sides by 8:
[tex]a=-\frac{4}{8}=-\frac{1}{2}[/tex]Now we replace the values of "a" and "c" in equation (1):
[tex]4(-\frac{1}{2})-2b-3=-13[/tex]Simplifying:
[tex]\begin{gathered} -2-2b-3=-13 \\ -2b-5=-13 \end{gathered}[/tex]Adding 5 to both sides:
[tex]\begin{gathered} -2b=-13+5 \\ -2b=-8 \end{gathered}[/tex]Dividing both sides by -2:
[tex]b=-\frac{8}{-2}=4[/tex]Therefore, the values of the constants are:
[tex]a=-\frac{1}{2},b=4,c=-3[/tex]