y=a(x-h)^2+kvertex is (h,k)and when a is positive it opts up svertex at -2,5 y=a(x+2)^2+5if we expand and set to zero0=ax²+4ax+4a+5dunnow how to solveI do know that A is positivealso since we gotvertex is (-2,5)it opens upthe vertex is the lowest point the zeros are where y=0notice lowest point is (-2,5)5 is the lowest point5>0, so there are no zeroes no solutions for that 2nd equation
