Answer:
The value of a is;
[tex]a=5[/tex]Explanation:
Given a quadratic equation;
[tex]y=ax^2+bx+c[/tex]The values of x and y are given in the table.
to get the value of c,
at x =0, y=2.
let us substitute into the equation;
[tex]\begin{gathered} y=ax^2+bx+c \\ 2=a(0)^2+b(0)+c \\ 2=c \\ c=2 \end{gathered}[/tex]Since, we have the value of c. the equation becomes;
[tex]y=ax^2+bx+2[/tex]let us proceed to find a and b.
at x=-1, y=16
also at x=1,y=-2
let us substitute the two values;
[tex]\begin{gathered} atx=-1,y=16 \\ y=ax^2+bx+2 \\ 16=a(-1)^2+b(-1)+2 \\ 16=a-b+2 \\ a-b=16-2 \\ a-b=14\text{ -------1} \\ at\text{ x=1, y=-2} \\ y=ax^2+bx+2 \\ -2=a(1)^2+b(1)+2 \\ -2=a+b+2 \\ a+b=-2-2 \\ a+b=-4\text{ ------2} \end{gathered}[/tex]To get a add equation 1 and 2 together;
[tex]\begin{gathered} a-b+a+b=14-4 \\ a+a-b+b=10 \\ 2a=10 \\ a=\frac{10}{2} \\ a=5 \end{gathered}[/tex]Therefore, the value of a is;
[tex]a=5[/tex]
