find the vertex-
To find the vertex of the equation ax^2 + bx + c the x-value of the vertex is -b/2a and you find the y-value by plugging in whatever number you get into the equation and solving
so like 6/3=2^2 thats all i got so far?
You can just memorize that h = –b / (2a) and then plug your value for "h" back into "y =" to calculate "k". If you're allowed to use this formula, you can then more quickly find the vertex, because simply calculating h = –b / (2a) and then finding k is a lot faster than completing the square. Hope this helps!
no y = ax^2 + bx + c you have y = 3x^2 + 6x - 4 a is 3 b is 6 c is -4 the x-coordinate of the vertex is \(\dfrac{-b}{2a}\)
-6/2(3)? so -1 right?
I think so!
Now you can plug that x-value into the equation to find the y-value at that x-value y = 3x^2 + 6x - 4 y = 3(-1)^2 + 6(-1) - 4 y = ??
um 3-10?
or 3-(-10)?
I believe it is 3-(-10)
waittt its (1,-7) right?
one part at a time 3 - 6 - 4 3 - 10 so yeah it's -7!
thanks bro B)
u really helped
It was my pleasure!
wait where did the 10 come from?
I only did it that way because it was the way you were doing it haha 3 - 6 - 4 -6 -4 = -10 so 3 - 10
ohhh I see thanks bro
otherwise 3 - 6 - 4 -3 - 4 = - 7 Of course, no problem!
so basically like b a c?
what is b a c
3 - 6 - 4 = 3 + (-6) + (-4) = 3 + (-6 + (-4)) = 3 + (-6 - 4) = 3 + (-10) = 3 - 10 = -7 if that makes it easier to see
okay yeah thats better thanks
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