Question

For each quadratic equation, find the:

a.) Roots
b.) y-intercept
c.) Vertex

y= 4(x-4)^2 +16

Answer 1

Ok so for the Roots I got (2,6) is that correct?
Also, I got 80 for the y-intercept. Is that also correct?

Answer 2

I'm having a bit hard time finding out the vertex, so I also need help on that.

Answer 3

for y-intercept plug in x=0. \[y= 4(0-4)^2 +16 = 4(-4)^2+16 = 80\] the y-intercept is correct

Answer 4

Yea I thought so to :) Thanks!

Answer 5

to find the roots you solve for your x-values, \[4(x-4)^2=16\]\[(x-4)^2 = 4\]\[\sqrt{(x-4)^2}= \sqrt{4}\]\[x-4 =2\] \[x=6\]

Answer 6

plug x back into the equation to find y

Answer 7

Yup so you get (2,6)

Answer 8

I'm just having a hard time finding the vertex.

Answer 9

oh wait, i did something wrong :( i think you do this: \[(x-4)^2 = 4\] \[x-4 =-2\] and \[x-4=2\] so yes you get 2 and 6 and tehn for the vertex, let me see...

Answer 10

Yup. Ok I just need help figuring the vertex. I don't just want the answer, I wan't to know how to do it too, u no?

Answer 11

Mmhmm. First you need to find the Axis of symmetry, the line that cuts the parabola in half. And the point on the graph that is represented by the x and y-value. In order to find it,you have the formula \[x= \frac{ -b }{ 2a }\] so you have to expand the formula \[y=4(x-4)^2+16\]

Answer 12

So expand it and tell me what you get :)

Answer 13

Ok so I usually do this in my calculator but it's messed up so I'm gunna do it your way :) Ok so were using the Quadratic Formula right?

Answer 14

\[y=4(x^2-8x+16)+16 = 4x^2-32x+64+16 =4x^2-32x+80\]

Answer 15

It's a bit confusing...

Answer 16

So now what you have to do is apply \[x=\frac{ -b }{ 2a }\] with your "b" and "a" values from this equation we just expanded, and find your x-value. Once you find your x-value you plug that x-value back into\[y=4(x-4)^2+16\] to get your y-value The (x,) value you will derive will represent the coordinate on which the axis of symmetry your parabola lies on, which is your vertex. :)

Answer 17

I'm sorry but I'm seriously not getting it. It's ugghh a bit frustrating.

Answer 18

what are you not getting? Which portion precisely.

Answer 19

The whole x = -b/2a It looks like you didn't apply it to your equation^^^

Answer 20

What do you mean didn't apply it to my equation? i didn't write out my steps. I can write out my step if you like. Do you have the answer to what the vertex is?

Answer 21

Ok so I usually do this in my calc but its not responding. I never learned how to do this by hand yet so that's y I'm lost. :3

Answer 22

Oh, okay. Well just like you use the quadratic to find your values of x, (usually you get 2 vaules using a quadratic), you can use the equation \[x=\frac{ -b }{ 2a }\] to find THE X-VALUE of the vertex coordinate. once you find this value, you plug it back into your original (or expanded equation) to get THE Y-VALUE of the coordinate of the vertex

Answer 23

Also, there is this video if you don't understand my explanation :) http://www.virtualnerd.com/algebra-1/quadratic-equations-functions/graphing/graph-basics/vertex-example There, the quadratic expression had already been expanded, which i advised you to do.

Answer 24

Ok thanks for all your patience and work! I appreciate the help! :)

Answer 25

Does it make sense at all? :(

Answer 26

why expand? the equation y= 4(x-4)^2 +16 is already in vertex form

Answer 27

Not in the least. Sorry. But I'm gunna check that video out and ask my professor for one on one time. Thanks though!

Answer 28

there's a reason why it's called that

y= a(x-h)^2 +k

has the vertex (h,k)

Answer 29

ex:

y= 2(x-7)^2 +33

has the vertex (7,33)

Answer 30

Ohh... I was working AROUND that because i forgot about the (h,k)...

Answer 31

So the (4,16) is the vertex?

Answer 32

Yes.

Answer 33

Oh that's it?

Answer 34

Thanks Jim :)

Answer 35

np

Answer 36

That's it! But i tried explaining HOW to get the vertex as well... sorry or confusing you!

Answer 37

Well I didn't no it was that simple. My professor always makes things harder than they have to be. :)

Answer 38

I summon Jim when I cannot get a point across. He's a lifesaver.

Answer 39

Haha np girl. Your extremely smart by the way! Thanks for the help! :)

Answer 40

Oh thanks ^_^ and np!

Answer 41

:)

Answer 42

Thanks @jim_thompson5910 also! I appreciate the help guys! :)

Answer 43

yw