SOLUTION: Can you help me solve this please There are 3 equations in turning point form.(vertex form) y=a(x-0)^2+300 The points on the graph are (0,300)(-122.5,0) and (122.5,0) y=a(x-122.

Question 443087: Can you help me solve this please
There are 3 equations in turning point form.(vertex form)
y=a(x-0)^2+300 The points on the graph are (0,300)(-122.5,0) and (122.5,0)
y=a(x-122.5)^2+300 The points on the graph are (122.5,300)(245,0) and (0,0)
y=a(x+122.5)^2-300 The points on the graph are (-122.5,300) (-245,0) and (0,0)
Can you please show me how to find the "a".
i Know the answer is -0.02 for all of them but i cant figure out how?
Your help is appreciated.
thanks
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
There are 3 equations in turning point form.(vertex form)
y=a(x-0)^2+300 The points on the graph are (0,300)(-122.5,0) and (122.5,0)
Let x = 122.5 and y = 0 ; then solve for "a".
I get a = -0.02
---
y=a(x-122.5)^2+300 The points on the graph are (122.5,300)(245,0) and (0,0)
Let x = 0 and y = 0 ; then solve for "a".
I get a = -0.02
----
y=a(x+122.5)^2-300 The points on the graph are (-122.5,300) (-245,0) and (0,0)
Let x= 0 and y = 0 ; then solve for "a".
I get a = +0.02
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---
Cheers,
Stan H.
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