When h>0, the image of y=a(x-h)2 can be represented by parabola y=ax? ; Move the h unit in parallel to the right,

When h < 0, it is obtained by moving |h| units in parallel to the left.

When h>0, k>0 and parabola y=ax? Move h units in parallel to the right, and then move k units up, and you can get y=a(x-h)? +k image;

When h>0, k<0, move the parabola y=ax2 to the right by h units in parallel, and then move it down by | k units to get y=a(x-h)? +k image;

When h < 0, k >; 0, move the parabola to the left by |h| units in parallel, and then move it up by k units to get y=a(x-h)? +k image;

When h < 0, k<0, move the parabola to the left by |h| units in parallel, and then move it down by |k| units to get y=a(x-h)? The image of+K.

Parabolic vertex coordinate solution

Image of parabola y=ax2+bx+c(a≠0): when a >: 0, open "when a"