When x=0, we have y=3. This means we are looking for an equation with a constant of 3. This narrows down the options to choose from to (3) and (6).
(3) &y=x+ 3
(6) &y=x^2+ 3
The equation could be either linear or quadratic. Notice that the points are not arranged in ascending order and therefore, we have to plot them in a coordinate plane to investigate if the graph takes the form of a line or a parabola.
Since we can draw a straight line through the points, the equation that describes the table of values must be a linear equation. Therefore, (3) is the correct equation.
b Like in Part A, we will start by looking for a point where x=0.
The equation we are looking for has its y-intercept at the origin. This narrows down the options to choose from to (1) and (4).
(1) &y=x
(4) &y=x^2
Again, the values are not given in ascending order so to find out if its a quadratic or a linear function, we have to plot the points.
Since we can draw a parabola through the points, the equation that describes the table of values must be a quadratic equation. Therefore, (4) describes the values.
c Like in Parts A and B, we will start by looking for a point where x=0.
& cccccccccc
& && & & & & ↓ & & & &
& |c|c|c|c|c|c|c|
x & 3 & -2 & 1 & 0 & 2 & -3
y & 12 & 7 & 4 & 3 & 7 & 12
The equation we are looking for has its y-intercept in (0, 3). The only two options we have to choose from with this y-intercept is (3) and (6). Since we have already picked (3), the table of data must match with (6).
d Like in previous parts, we start by looking for a point with an x-coordinate of 0.
& cccccccccc
& && & & & & & & ↓ &
& |c|c|c|c|c|c|c|
x & -3 & 4 & 2 & -2 & 0 & -10
y & -10 & 11 & 5 & -7 & -1 & -31
The equation we are looking for has its y-intercept in (0, -1). There is only one such equation, namely (2).
