We can use points (-1,12), (0,10) and (1,12) to find the quadratic equation. First, we will work with the formula:
[tex]y=ax^2+bx+c[/tex]With the three points mentioned above, we make the proper substitution:
[tex]\begin{gathered} P_1=(-1,12) \\ \Rightarrow12=a(-1)^2+b(-1)+c \\ \Rightarrow12=a-b+c \\ P_2=(0,10) \\ \Rightarrow10=a(0)^2+b(0)+c \\ \Rightarrow10=c \\ P_3=(1,12) \\ \Rightarrow12=a(1)^2+b(1)+c \\ \Rightarrow12=a+b+c \end{gathered}[/tex]So, we have the following system of equations:
[tex]\begin{gathered} a-b+c=12 \\ a+b+c=12 \\ c=10 \end{gathered}[/tex]Now, substitute c=10 on the other equations to get the following:
[tex]\begin{gathered} c=10 \\ \Rightarrow a-b+10=12 \\ a+b+10=12 \\ \Rightarrow a-b=12-10=2 \\ a+b=12-10=2 \\ \Rightarrow a-b=2 \\ a+b=2 \end{gathered}[/tex]solving by elimination we get the value for 'a':
[tex]\begin{gathered} a-b=2 \\ a+b=2 \\ \Rightarrow2a=4_{} \\ \Rightarrow a=\frac{4}{2}=2 \\ a=2 \end{gathered}[/tex]we have that a=2 and c=10. We can now find the value for b:
[tex]\begin{gathered} a=2 \\ a+b=2 \\ \Rightarrow2+b=2 \\ \Rightarrow b=0 \end{gathered}[/tex]Now that we have that a=2,b=0 and c=10, we can write the quadratic function:
[tex]\begin{gathered} a=2 \\ b=0 \\ c=10 \\ y=ax^2+bx+c \\ \Rightarrow y=2x^2+0\cdot x+10 \\ y=2x^2+10 \end{gathered}[/tex]therefore, the quadratic function in standard form is y=2x^2+10
