I've been trying to learn about quadratic functions and taking one in standard form and putting it in vertex form. When I read the lesson explanation I think I understand it, but then when I do the practice problems the site has I always get them wrong, and I don't know what I'm not doing right. So I was wondering if I showed what I did step by step someone could tell me, where I'm going wrong.
The example problem is,
Put y = -2x[sup:3fvf8i11]2[/sup:3fvf8i11] - 5x - 3 in vertex form and give the coordinates of the vertex.
So this is what I did,
y = -2x[sup:3fvf8i11]2[/sup:3fvf8i11] - 5x - 3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x) -3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11] - -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) (-2)-5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
I then turned the middle middle part into a perfect square trinomial, -5/-2(2) = 5/4 so,
(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) = (x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11]
So then I had,
y = -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)-5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)25/16 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)25/16 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] -25/8 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] -49/8
Which makes the coordinates of the vertex,
(-5/4 , -49/8) which is wrong. The correct answer being, (-5/4 , 1 /8)
Now a positive 25/8, minus 3 would get me 1/8, so somewhere I must get mixed up over whether something is supposed to by a positive or a negative but, I can't see where, and I keep making the mistake everytime I work this problem out, or a problem like it.
Where am I messing up?
Sorry if the question was not very clear, or extremely dim. :? I find math very confusing.
The example problem is,
Put y = -2x[sup:3fvf8i11]2[/sup:3fvf8i11] - 5x - 3 in vertex form and give the coordinates of the vertex.
So this is what I did,
y = -2x[sup:3fvf8i11]2[/sup:3fvf8i11] - 5x - 3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x) -3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11] - -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
= -2(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) (-2)-5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
I then turned the middle middle part into a perfect square trinomial, -5/-2(2) = 5/4 so,
(x[sup:3fvf8i11]2[/sup:3fvf8i11] -(5/-2)x + -5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) = (x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11]
So then I had,
y = -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)-5/-2(2)[sup:3fvf8i11]2[/sup:3fvf8i11]) - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)25/16 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] (-2)25/16 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] -25/8 - 3
= -2(x + 5/4)[sup:3fvf8i11]2[/sup:3fvf8i11] -49/8
Which makes the coordinates of the vertex,
(-5/4 , -49/8) which is wrong. The correct answer being, (-5/4 , 1 /8)
Now a positive 25/8, minus 3 would get me 1/8, so somewhere I must get mixed up over whether something is supposed to by a positive or a negative but, I can't see where, and I keep making the mistake everytime I work this problem out, or a problem like it.
Where am I messing up?
Sorry if the question was not very clear, or extremely dim. :? I find math very confusing.