12 - Writing Quadratic Functions in Vertex Form - Part 1 (Graphing Parabolas)

More Lessons: www.MathAndScience.com
Twitter: / jasongibsonmath
In this lesson, you will learn how to write any quadratic function that is given in standard form into vertex form. Vertex form is superior in that we can sketch a graph of the function easily because the vertex can be read directly from the equation.
The method used to write quadratic functions in vertex form is to perform the completing the square operation on the standard form quadratic. We show this by solving numerous problems.

[@rangarajunanjundagowder4530]

Best best best teacher.His method of teaching is outstanding

[@juliobaez9721]

Jason is a tremendously articulate instructor.

[@tesfamariamworku6690]

I have seen many professors in many different courses, but you are on top of the list

[@georgettebeulah4427]

I can real understand what your saying I know how your explanation is and see where it is coming from thank you much

[@serenity6616]

You're just amazing. I'll never say thank you enough to you

[@Garrison86]

can you link the previous videos you are referring to in the video in you description section, that would be mucho appreciated, thank you so much for the comprehendible lessons.

[@arriarriarri]

The first math teacher i've seen with beautiful handwriting :_(

[@moontuple]

Lol

[@nigihepeace]

Superior teacher!!!!! Nice!!

[@joumanaswailem6245]

Thank you so much for this amazing explanation, BTW there's an easier way to solve the 1st eg. And it is X2 +2x-3 X. 3 X. -1 We separate the x square And the multiplication of C , and that should be the total of bx by doing (×+3)(×-1)=0 so it'll be ×=-3 & ×=1 I wish that you understood

[@elizabethsakine9439]

This has helped me so much

[@carmelpule8493]

The drawing of this parabola y - k = a( x - h)^2 may be looked upon as ...... Two straight lines (y2 = (x -- h ) multiplied together, then multiplied ( amplified) by a third line y3 = (a) then the answer added ( shifted) to the straight line y4 = k where k, h will be the vertex. Looking at a parabola as multiplying two straight lines, then amplifying them, then shifting the products up and down is also a useful way to draw a parabola.

[@ynamanalo9900]

can you teach us about how you solve quadratic equation word problems?

[@vivicheung4843]

Thank you so much your hard working

[@lostpineschaptertexasmaste3773]

It makes sense that, if you start with an equation in standard form, you need to convert it to vertex form to get the vertex. That part of the video is great. It does NOT make sense to THEN go to the trouble of "foiling" the vertex form back out in order to find the x-intercepts. "Foiling" the vertex form back out ALWAYS takes you right back to the standard form of the equation. In other words (unless you make an error somewhere along the way) you ALWAYS wind up right back with the standard equation you started with. Why do all that work (which takes up probably 5 to 8 minutes of this video) AND take the risk that you'll make an error somewhere? nstead, just pop back to your standard form, set y = 0, and see if you can factor the resulting equation to get the x-intercepts... IF you really need to find the x-intercepts. Often though, you don't need to. There is another piece of information one can glean from the standard form of a quadratic equation that Mr. Gibson never mentions. In addition to the value of "a" telling you whether the parabola opens upward or downward, the value "c" tells you the y-intercept. Once you know the vertex, and the y-intercept, you can do a quick sketch of the parabola without ever finding the x-intercepts... and you may discover that your parabola doesn't even have any x-intercepts. And, if x-intercepts exist, from your quick graph, you can usually get a good enough "estimate" of where they are to pick out the correct answers on a multiple choice test. The quadratic formula is a great tool in real life, and it is often useful on teacher-made tests in class. But if you find yourself having to apply the quadratic equation in the middle of a timed SAT-exam, there is a good chance you've overlooked a simpler method of getting to the solution. The quadratic formula just takes too long and it's too easy to make a calculation error. Just my two cents.

[@romanemul1]

How about posting some making of video how you do your stuff ?

[@tavia6265]

this is great info but how would i do f(x)x^2-3x ?

[@iramishra6343]

Awesome!!!!

[@shwinside]

thank you so much!!

[@chandamandona1940]

Thankyou ❤

[@lloyds.2420]

how do we know a positive constant leads to a parabola opening upwards?

[@e4va444]

You can test it out but a positive leading coefficient will always have the parabola opening upwards, you can test it on desmos or other sites. Similarly a negative leading coefficient will always have the parabola openings downwards.

[@angelayj3272]

This is still so confusing for me. How is the vertex (-3,4) is this equation?: y=2x²+12x+22

[@ellaknickerbocker7821]

thanks, pray i do good on my math exam 🙏

[@dorkle9085]

What if I have it in the form of y = ax^2 + b?

[@vespa2860]

Do you mean bx?

[@dorkle9085]

@@vespa2860 nope, just B

[@dorkle9085]

Also nevermind I figured it out thank you

[@racquelsabesaje4562]

y-1+2=2(x2-2x+1)

[@Maggie-wz6ez]

f(x)=-2x^2 +7x would i just add a zero as a placeholder for the c term? can anyone help me out I have an exam tomorrow.

[@MathAndScience]

Yes, c=0 in this case because it isn't there. Well, it is really there it is just that "c" is equal to zero!

[@racquelsabesaje4562]

y+1=2(x2-2x+1)

[@racquelsabesaje4562]

f(x)=ax2+bx+c

[@racquelsabesaje4562]

1=2x2-4x+2

[@racquelsabesaje4562]

y+1=2(x2-2x=1)

[@racquelsabesaje4562]

y=x2+2x-3

[@racquelsabesaje4562]

f(x)=2x2-4x+1

[@racquelsabesaje4562]

0=x2+2x-3

[@racquelsabesaje4562]

f(x)=x2+2x-3

[@racquelsabesaje4562]

y=2x2-4x+1

[@racquelsabesaje4562]

y+1=2(x-1)2

[@racquelsabesaje4562]

y+3(2/2)2=x2+2x+(2/2)2

[@racquelsabesaje4562]

y-1=2(x2-2x)

[@racquelsabesaje4562]

4=x2+2x+1

[@racquelsabesaje4562]

y-1=2x2-4x

[@stellaoffei6114]

Wow

[@racquelsabesaje4562]

y+3+(2/2)2=x2+2x+(2/2)2

[@racquelsabesaje4562]

0=x2+2x+1-4

[@Micana1980]

Tanx

[@racquelsabesaje4562]

y+3=x2+2x

[@racquelsabesaje4562]

y+4=(x+1)2

[@racquelsabesaje4562]

quadratic

[@racquelsabesaje4562]

0+1=2(x-1)2

[@racquelsabesaje4562]

y+1=2(x-1)(x-1)

[@racquelsabesaje4562]

y+4=x2+2x+1

[@racquelsabesaje4562]

y+4=x2=(x)(x)

[@racquelsabesaje4562]

y+4=(x +1 )(x + 1 )

[@racquelsabesaje4562]

(x-1)(x+3)=0

[@racquelsabesaje4562]

y+4=x2+1x+1x+1

[@racquelsabesaje4562]

x=1 x= -3

[@Dereck27]

Not understanding how you’re getting the vertex at all. So many steps. How can a person be expected to memorize this crap? Man this is frustrating.

[@jaykumarjoshi5173]

You mention to drafting the equation but do show and therefore concept is not clear to my mind only empty words are not enough l do not understand anything of your vidio sorry

[@racquelsabesaje4562]

y+3+1=x2+2x+1

[@racquelsabesaje4562]

0+4=(x+1)2

[@racquelsabesaje4562]

y+4=(x+1)(x+1)