The graph of y=ax²+c has its vertex at (0,4) and passes through the point (3,-5). Find the value of a.?
回答 (2)
2020-06-07 4:10 pm
✔ 最佳答案
The vertex of the graph has its vertex at (0, 4).Hence, the equation of the graph is: y = ax² + 4
The graph passes through the point (3, -5).
Hence, plug x = 3 and y = -5 into the equation:
(-5) = a(3)² + 4
9a + 4 = -5
9a = -9
a = -1
2020-06-07 4:49 pm
If the vertex is at (0, 4) we have:
4 = a(0)² + c
so, c = 4
Hence, y = ax² + 4
As it passes through (3, -5) we have:
-5 = a(3)² + 4
so, a = -1
Therefore, y = -x² + 4 or 4 - x²
:)>
4 = a(0)² + c
so, c = 4
Hence, y = ax² + 4
As it passes through (3, -5) we have:
-5 = a(3)² + 4
so, a = -1
Therefore, y = -x² + 4 or 4 - x²
:)>
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