Question

Could use some help asap,
Image includes Rewriting a quadratic function in standard form... and more <3 mainly need help with the rewriting...

Answer 1

http://imgur.com/wEWg7ZL

Answer 2

@Zale101 if u got time ^_^

Answer 3

What do you know about solving x or y intercepts?

Answer 4

What do you do to the equation to make it possible to solve for the desired variable (e.g. x or y.) ?

Answer 5

Currently we'd need to complete the square in order to put it in standard form right? that way i can determ the x and y intercepts and the vertex, correct?

Answer 6

To solve for the y -intercept, your desired variable is y. To solve for y, you set x=0 to get the y-intercept.

Answer 7

Do that and we will discuss on how to solve for the x-intercept later.

Answer 8

y=-5

Answer 9

Yes. So when x=0, y=-5
So the y-intercept point is at (0,-5)

Now, to solve for the x-intercept, you set y=0. Now, solving for intercept is not that simple as fr solving for the y-intercept. This time, you either factor it or use the quadratic formula. In this case, x has two solutions. Are you good at factoring or plug and chug to the qudratic formula \(x=\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 10

Clear?

Answer 11

yup, the main issue i have is with getting it into standard form ;-;

Answer 12

"solving for the x intercept is not that simple as for solving for the y-intercept. "

Answer 13

Well, your quadratic is already in standard form. Standard form is when quadratics are written as \(ax^2+bx+c=0\) or \(ax^2+bx+c=y\).

I think what you meant was how to convert a standard equation to the vertex form \(y=a(x-h)^2+k\), where h,k is the vertex.

Answer 14

Yeah, forgive me ^_^ Meant vertex form.. book states it as "Standard Form".

Answer 15

Oh. Alright. I am assuming you want to convert the equation into the vertex form so the vertex can be easily manipulated and be visually seen on a graph. Is that correct? If so, you can complete the square to get there OR use the vertex formula \((\frac{-b}{2a},f(\frac{-b}{2a}))\)

Answer 16

Yeah, that's pretty much what i need to do ^_^ could you explain how to complete the square? as i am not 100% sure on how to do it

Answer 17

you havent been typing for 20 min have you? @Zale101 lol <3

Answer 18

Sure. Here is an example of how to complete the square.

let us say you have a quadratic that happens to be like this:
\(y=2x^2+12x-4\)
First, you place a parenthesis around any terms multiplied to x.
\(y=(2x^2+12x)-4\)
Second, you factor out the common factor. In this case, our common factor is 2.
\(y=2(x^2+6x)-4\)
Secondly, you then proceed with completing the square. To do that, you take half of 6 and square it. In this case, our coefficient is 6x.
\((\frac{6}{2})^2=3^2=9\)
\(y=2(x^2+6x+9-9)-4\)
kick the -9 out of the parenthesis, BUT to do that, -9 cannot leave the parenthesis only if it gets multiplied by 2. Hence the 2 that was factored out.
\(y=2(x^2+6x+9)-2(9)-4\)
\(y=2(x^2+6x+9)-18-4\)
\(y=2(x^2+6x+9)-18-4\)
factor further

\(y=2(x+3)(x+3)-18-4\)

\(y=2(x+3)^2-22\) <--- Final Answer

Answer 19

Sorry, i was kinda busy

Answer 20

no problem :)

Answer 21

Follow the steps i gave above to your given quadratic.

Answer 22

will do now.

Answer 23

Hey zale, where'd you get +9-9?
i understand here +9 fits in because (b/2)^2 but the -9 i dotn quite understand

Answer 24

After i completed the square, i had 6/2=3 and then 3^2=9. I can't add 9 to the mix when i do not have its same number subtracted to it. If i just add 9 by iteself, then it is not longer the same equation.
For instance y=x^2+12 is that same as y=x^2+12+9-9, at the end 9-9 will be zero. y=x^2+12+0 which then will equal to the same original equation y=x^2+1.

Answer 25

i see!!!
Thank you! :D

Answer 26

but if i just add 9 by itself y=x^2+12+9 then y=x^2+21 does NOT equal to equation y=x^2+12. These are two different equations.

Answer 27

No Problem.