Asked by natalie
I cannot for the life of me remember how to do this- Find the vertex of the graphs of the functions:
Function #1: y=(x-4)(x+2)
AND
Function #2: y=2x^2-4x+1
Can someone help me get it PLz!
OK - to find the vertex of these functions you must have them in quadratic form where the first coefficient in front of the x^2 is a, the linear term is b, and the constant is c. You then do -b/2a. For the first one, you must FOIL it out to get x^2-2x-8. b is -2 and a is 1. So, the x-coordinate of the vertex is -(-2)/2(1) which is 1. Then plug it back in for x so y=(1-4)(1+2), y=-9. The vertex is (1,-9). Do the same for the second function. You should end up with (1,1); if you don't get this please post again.
Function #1: y=(x-4)(x+2)
AND
Function #2: y=2x^2-4x+1
Can someone help me get it PLz!
OK - to find the vertex of these functions you must have them in quadratic form where the first coefficient in front of the x^2 is a, the linear term is b, and the constant is c. You then do -b/2a. For the first one, you must FOIL it out to get x^2-2x-8. b is -2 and a is 1. So, the x-coordinate of the vertex is -(-2)/2(1) which is 1. Then plug it back in for x so y=(1-4)(1+2), y=-9. The vertex is (1,-9). Do the same for the second function. You should end up with (1,1); if you don't get this please post again.
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