Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(1,-4)\\(3,-16)\\(7,14)$
Substitute: $-4=a(1)^2+b(1)+c\\-16=a(3)^2+b(3)+c\\14=a(7)^2+b(7)+c$
We have the system: $a+b+c=-4\\9a-3b+c=-16\\49a+7b+c=14$
Subtract the second equation from the third equation:
$40a+10b=30\\
\rightarrow 4a+b=3$ (1)
Subtract the first equation from the second equation:
$8a-4b=-12\\
\rightarrow 2a-b=-3$ (2)
Add equation (1) to equation (2):
$6a=0=-4\\
\rightarrow a=0$
Substitute $a$ to equation (2):
$2(0)-b=-3\\
\rightarrow b=3$
Find $c$:
$0+3+c=-2\\
\rightarrow c=-7$
Hence, $a=0\\b=3\\c=-7$
Substitute back to the initial equation: $y=3x-7$
The given points satisfy the same linear function. Hence, they lie on a straight line.
