Vertex Of A Quadratic Function In Standard Form

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L3

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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.
 
L3 said:
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

Put y = -2x[sup:3c676biy]2[/sup:3c676biy] - 5x - 3 in vertex form and give the coordinates of the vertex.

So this is what I did,

y = -2x[sup:3c676biy]2[/sup:3c676biy] - 5x - 3 >>> Why not just use h = -b / (2a) here?
= -2(x[sup:3c676biy]2[/sup:3c676biy] -(5/-2)x) -3

...

(-5/4 , -49/8) which is wrong. The correct answer being, (-5/4 , 1 /8)
Click to expand...

This will give you h = -5/4, which is the x coordinate for the vertex. Now, plug that back into your original equation and find k.

This is an exercise in converting y = ax^2 + bx + c to the form y = a(x+h)^2 + k.

Note that a is the same in both forms for the same function.

Therefore, plugging the x coordinate at the vertex back into the original problem, we can find the y coordinate of that vertex. This will also give us k for finding the a(x+h)^2 + k form of the function.

\(\displaystyle k = -2 (\frac{5}{-4})^2 + \frac{25}{4} - 3\)

And we don't have as much opportunity for a mistake. The vertex is at (h, k).
 
Yes, but the site I was doing it on taught both ways, and I sort of wanted to be able to do both ways. Also, now I'm just curious as to what I did wrong.
 
Let me give you the general formula for completing the square.

\(\displaystyle a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant term}}\)

Enter in your values of a=-2, b=-5, c=-3 and you have it.

\(\displaystyle -2(x+\frac{5}{4})^{2}+\frac{1}{8}\)

BTW, I think you just have an arithmetic problem. -25/8+3=-1/8

Though, I did not look over it in depth.

You could do it like so:

\(\displaystyle -2x^{2}-5x=3\)

\(\displaystyle -2(x^{2}+\frac{5}{2}x+\frac{25}{16})=3-\frac{25}{8}\)
\(\displaystyle .......................\uparrow{(\frac{5}{4})^{2}}......\uparrow{-2(\frac{25}{16})}\)

\(\displaystyle -2(x+\frac{5}{4})^{2}=\frac{-1}{8}\)

\(\displaystyle -2(x+\frac{5}{4})^{2}+\frac{1}{8}\)
 
L3 said:
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:3jg29tnp]2[/sup:3jg29tnp] - 5x - 3 in vertex form and give the coordinates of the vertex.

So this is what I did,

y = -2x[sup:3jg29tnp]2[/sup:3jg29tnp] - 5x - 3
= -2(x[sup:3jg29tnp]2[/sup:3jg29tnp] -(5/-2)x) -3
= -2(x[sup:3jg29tnp]2[/sup:3jg29tnp] -(5/-2)x + -5/-2(2)[sup:3jg29tnp]2[/sup:3jg29tnp] - -5/-2(2)[sup:3jg29tnp]2[/sup:3jg29tnp] ) - 3 <<< That does not look correct


= -2(x[sup:3jg29tnp]2[/sup:3jg29tnp] +(5/2)x + 25/16 - 25/16) - 3

= -2(x[sup:3jg29tnp]2[/sup:3jg29tnp] +(5/2)x + 25/16) + 25/8 - 3


I then turned the middle middle part into a perfect square trinomial, -5/-2(2) = 5/4 so,
(x[sup:3jg29tnp]2[/sup:3jg29tnp] -(5/-2)x + -5/-2(2)[sup:3jg29tnp]2[/sup:3jg29tnp]) = (x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp]

So then I had,
y = -2(x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp] (-2)-5/-2(2)[sup:3jg29tnp]2[/sup:3jg29tnp]) - 3
= -2(x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp] (-2)25/16 - 3
= -2(x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp] (-2)25/16 - 3
= -2(x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp] -25/8 - 3
= -2(x + 5/4)[sup:3jg29tnp]2[/sup:3jg29tnp] -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.
Click to expand...
 
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