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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
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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
8.9k points