9.4k views
4 votes
20 POINTS!!!!!!!

The graph of y = ax2 + bx + c is a parabola that opens up and has a vertex at (-2, 5). What is the solution set of the related equation 0 = ax2 + bx + c?

Ø
{-2}
{5}

User Crooked
by
7.3k points

2 Answers

2 votes

Answer: your correct answer would be no solution

Explanation:

User Gilad Green
by
8.1k points
2 votes
Hello,



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

User Renan Bandeira
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories