y+2=x (Red line) Adding a +2 to equation by the y side, changes that the equation is not written in standard form, once rearranged the same equation of y=x-2 is obtained which shifts the graph to cut the y-intercept …show more content…
The last set of equations for the parabola is the inclusion of both the p and q values. y=x^2 (Blue line). Turning point at the origin (0,0) y-1=〖(x-2)〗^2 which can also be written as y=〖(x-2)〗^2+1. (Green line) Thus the turning point is (2,1). This makes the graph shift both right of the y-axis and upwards. y+1=〖(x+2)〗^2 which can also be written as y=〖(x-2)〗^2-1. (Red line) Thus the turning point is (-2,-1). This makes the graph shift both left of the y-axis and downwards.
This set of line is that of a hyperbola. The standard form of a hyperbola is y=a/(x-p)+q where if a>0 the hyperbola will be in quadrant 1 and 3. If a1 then the function is increasing. In the equations below we can see that the p-value shifts the graph either left or right. The asymptote is the line y=q. Thus q shifts the graph up or down. To determine the reference point, which shows that the graph has shifted we can use the co-ordinates (p;a+q)
y=2^x (Blue) The asymptote is y=0 and the reference point is (0;1) y=2^(x-2) (Red) The asymptote is y=0 and the reference point has moved to (2;1) y=2^(x+2) (Green) The asymptote is y=0 and the reference point has moved to