Question

What is the axis of symmetry of the parabola represented by the function y=2x2 + 12x + 10?

Answer 1

Graph it.

Answer 2

It's going to be a parabola.

Answer 3

Here this may help=) http://www.mathwarehouse.com/geometry/parabola/axis-of-symmetry.php

Answer 4

I dont know how to graph equations

Answer 5

You can do it one of two ways. You can either graph it by using a graphing calculator or you can make a table.

Answer 6

i dont know how to do it like that, my calculator doesnt have an x, does it?

Answer 7

just use

for a parabola

\[y = ax^2 + bx + c\]

the line of symmetry is

\[x = \frac{-b}{2a}\]

Answer 8

in your question b = 12 and a = 2

Answer 9

a 3rd method is to factor the equation and the line of symmetry is half way between the zeros

Answer 10

Campbell_st how did you obtain the equation x=-b/2a did you solve for x?

Answer 11

Oh my gosh, im so sorry but i'm really confused :/

Answer 12

its just a know fact.... the general quadratic formula gives the zeros...

and x = -b/2a is the midpoint between the zeros...

I learnt it is year 10

Answer 13

It has been quite awhile for me as well.. The formula looks familiar from a previous class...

Answer 14

ok... so use the equation

for any parabola

\[y =ax^2 + bx + c\]

the line of symmetry can be found using

\[x = \frac{-b}{2a}\]

so in your question, a = 2 and b = 12

so its

\[x = \frac{-12}{2 \times 2}\]

so the line of symmetry is x = -3...

which makes sense as the zeros are

x = -5 and x = -2

Answer 15

Mind if I butt in @AmberLynn127 The axis of symmetry is the y axis, because the function is even. Even functions are symmetric about the y axis.

Answer 16

and the reason it looks familiar is that

the general quadratic formula is

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

take out the radical and you have

\[x = \frac{-b}{2a}\]

Answer 17

i know that! i just dont know how to make the graph, if i had the graph i could figure it out

Answer 18

So @campbell_st do you obtain x=-b/2a by deriving the equation y=ax^3+bx+c

Answer 19

the function isn't even

f(-x) = 2(-x)^2 + 12(-x) + 10

which is

f(x) = 2x^2 - 12x + 10

so f(-x) does not equal f(x)

the function is neither odd nor even

Answer 20

@AmberLynn127 That's good for you. There is the procedue to graph quadratic functions
Find x intercepts
Find y intercepts
Find the co ordinates of the vertex.

Answer 21

hey wait!! so i have another one, and i dont remember how to set up the problem to solve it, so any help?
Michelle has a garden in the shape of a square. How long is one side of the square garden if it has an area of 144 ft2?

Answer 22

well for any parabola...

y= ax^2 + bx + c

x = -b/2a will always give the line of symmetry

then if the curve is

\[x = ay^2 + by + c\]

the line of symmetry is

\[y = \frac{-b}{2a}\]

Answer 23

@campbell_st I didn't realize that the function is not even.

Answer 24

Okay thanks @campbell_st

Answer 25

if you look at it, it can't be symmetry about the y axis...

is it was then the parabola would be the difference of 2 squares or

y = ax^2

Answer 26

well in a square all sides are equal so then

\[144 = l^2\]

where l is the side length

take the square root and use the positive answer..

you can't have a negative side length

Answer 27

so the square root of 144?

Answer 28

The axis of symmetry is the vertical line through x=-b/(2a), which in this case is not x=0 (the y-axis). x=-b/(2a) is also the coordinate of the vertex. As Campbell has correctly pointed out, this x coordinate is x=-12/(2*2) = -3. To find the vertex, substitute this x value (-3) into the original quadratic and calculate the y value. Then you'll have the coordinates of the vertex.

Answer 29

thats correct

Answer 30

Amber-Lynn: mind summarizing now what you know and what you still need to figure out?

Answer 31

@mathmale thats provided its not a parabola in the form

y = ay^2 + by + c

Answer 32

Its just hard for me to remember every thing because school is CRAZY and want us to memorize every single equasion form in a week!

Answer 33

Campbell? You don't really want y on both sides of that equation, do you? Unsure of what you're trying to say.

Answer 34

Amber-L: I sympathize. But I do still suggest you read over the discussion you and several others have had about your original question and then list what you've learned and what you still need to learn to answer that question.

Answer 35

i have not learned a thing. It didn't help. They didn't explain anything

Answer 36

At least you know even and odd functions.

Answer 37

Is everyone talking to me? or each other? cause i thought you guys were talking to each other,,

Answer 38

Actually I referred to you @AmberLynn127

Answer 39

IM LOST

Answer 40

A-L: Sorry you feel lost. Several people have invested considerable time and thought into helping you address the problem you need to solve. I was asking that you make the effort to skim through what they have written and summarize their input for your own understanding of this problem. Were you able to do this, you'd be in a correspondingly better position to explain what it is that you still don't understand. Try to put yourself into the position of those who tried to help you, only to be told by you, "i have not learned a thing. It didn't help. They didn't explain anything." I won't be adding any more to this discussion. Good luck.

Answer 41

Umm actually i read everything, but it seemed like they were on whole nother topic.

Answer 42

Perhaps you need to read this material again, critically, deciding for yourself what's helpful and what's not. Others can do only so much for you; at some point you yourself have to step in, get involved, spend the time and effort, and take control.