Hi Jason, I am very glad, you presented this topic. We exchanged views on this a few years ago, when I told you, that to go from general form to vertex form was a very important lesson in german middle and high schools. Extremely well explained! Congratulations and Cheers, Rolf
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.
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.
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
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.
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.
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
Best best best teacher.His method of teaching is outstanding
Hi Jason, I am very glad, you presented this topic. We exchanged views on this a few years ago, when I told you, that to go from general form to vertex form was a very important lesson in german middle and high schools. Extremely well explained! Congratulations and Cheers, Rolf
I have seen many professors in many different courses, but you are on top of the list
You're just amazing. I'll never say thank you enough to you
Jason is a tremendously articulate instructor.
I can real understand what your saying I know how your explanation is and see where it is coming from thank you much
The first math teacher i've seen with beautiful handwriting :_(
Lol
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.
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.
Superior teacher!!!!! Nice!!
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
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.
This has helped me so much
can you teach us about how you solve quadratic equation word problems?
this is great info but how would i do f(x)x^2-3x ?
thanks, pray i do good on my math exam 🙏
How about posting some making of video how you do your stuff ?
how do we know a positive constant leads to a parabola opening upwards?
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.
Thank you so much your hard working
This is still so confusing for me. How is the vertex (-3,4) is this equation?: y=2x²+12x+22
What if I have it in the form of y = ax^2 + b?
Do you mean bx?
@@vespa2860 nope, just B
Also nevermind I figured it out thank you
Thankyou ❤
y+3(2/2)2=x2+2x+(2/2)2
y-1+2=2(x2-2x+1)
y-1=2(x2-2x)
thank you so much!!
y+1=2(x2-2x=1)
0=x2+2x+1-4
f(x)=ax2+bx+c
f(x)=x2+2x-3
4=x2+2x+1
Awesome!!!!
y=x2+2x-3
y+3=x2+2x
y+3+1=x2+2x+1
y+4=x2=(x)(x)
y+4=(x +1 )(x + 1 )
y+4=(x+1)2
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.
y+4=x2+1x+1x+1
(x-1)(x+3)=0
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.
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!
quadratic
Tanx
Wow
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
y+3+(2/2)2=x2+2x+(2/2)2
f(x)=2x2-4x+1
0=x2+2x-3
y-1=2x2-4x
y=2x2-4x+1
y+1=2(x-1)(x-1)
y+1=2(x-1)2
y+4=(x+1)(x+1)
0+1=2(x-1)2
0+4=(x+1)2
x=1 x= -3
y+1=2(x2-2x+1)
1=2x2-4x+2
y+4=x2+2x+1