Answer:
quadratic with a common second difference of 4
Step-by-step explanation:
quadratic equations: Quadratic equations are the polynomial equations of degree 2 example - a[tex]x^2\\[/tex]+ b[tex]x[/tex] + c = 0.
[tex]\begin{array}{cc}x&y\\-2&8\\-1&2\\0&0\\1&2\\2&8\end{array}\right][/tex]
from the table it is clear that,
function is y = 2[tex]x^2\\[/tex]
so the function is not linear as the equation specifies.
formula to find second difference of a quadratic equation:
f(x + 2) - 2f(x + 1) + f(x)
therefore;
[tex]2(x+2)^2[/tex] - [tex]4(x+1)^2[/tex] + [tex]2x^2[/tex]
[tex]2(x^2 + 4x + 4) - 4(x^2 + 2x+ 1 ) + 2x^2[/tex]
[tex]2x^2 + 8x + 8 - 4x^2 - 8x - 4 + 2x^2[/tex]
[tex]4x^2 - 4x^2 + 8x - 8x +8 - 4[/tex]
4
therefore, the answer is quadratic with a common second difference of 4
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