I've never understood graph transformations properly so seeing graphs drawn helps a lot. Thank you😂

@Alfakkin

6 ай бұрын

Me too

@mauriciocastillo7302

6 ай бұрын

It's not too complicated to understand: 1) in y=√x, when we plug in x values such as 0, 4, 9, 16, 25, 36, 49 (perfect squares), we notice the y value steadily increases, but the gaps in the x value get bigger and bigger, hence it looks like the function almost flattens out. 2) in y=-√x, we notice it's just -1 times the original function, so the graph is the same but with negative y values, hence the reflection over the x axis. 3) in y=√-x, in regards to real numbers, we notice this is only true with negative x values, so it will be the original function except going in the negative direction. 4) in y=√(x-2), the trickiest one, we notice we need a bigger x value to get the same y value from the original function: for example, to get the same in y=√(x-2), we need (x+2). Say we had 9 in the original, we would need 11 in this function to get the same y value of 3. 5)in y=√(x) + 2, we notice it's just the original function but with + 2 added to every answer, meaning we offset the function 2 units up as its starting place (and every other value) is 2 units greater.

Thank you very much dear Sir for these wonderful videos and for using Chalk and board Very tired of teachers using digital board We must preserve the tradition God bless ❤️😊

Looking jacked Mr H

This is powerful and not nearly taught enough

Where have you been all my Life... I'm 46 and my spouse and I have a 14 year old son and your videos have been a life saver! Thank you so very much

@mrhtutoring

6 ай бұрын

I hope the channel will help your 14 year old also.

Thats some triceps there 😮

@josephkeres4604

6 ай бұрын

That’s the point of the shirt,bro.

@_Facade

6 ай бұрын

@@josephkeres4604 😁

Thank you!!!!! ❤🙏

Transformations on reciprocal graphs with an oblique asymptote always confused me in my college algebra class.

“And if you graph x^2, you get what looks like me flexing with my guns”

Teh 3rd one was the only one I didn't know about

Amazing

The half sideways parabola does prove that extraneous solutions exist.

Muito bom. Thanks to you.

Man, i remember how i struggled with these in highschool. XD

ขอบคุณครับ

Sir im not understand , the graph of root under -x. How can you draw if it is imaginary

@ninolatimer1863

6 ай бұрын

For sqrt(-x), if x = -1, it becomes sqrt(- -1), negatives cancel out making sqrt(1) and same for any other negative value

@luisclementeortegasegovia8603

6 ай бұрын

If √(- x) then for negative - x we have √(-(- x)) = √x

@jasonrubik

6 ай бұрын

When X is positive then you would get an imaginary number. He is only showing the real cartesian plane. Thus the imaginary results are hidden. This is true for every single equation on the chalkboard here.

@DrDeuteron

6 ай бұрын

-(-x) = x

@mybeats7478

6 ай бұрын

Gentle man im hereby tell you that, This is a root function not modulus, it can't give real values for y it is undefined

Can u solve jee maths section question paper

Does a start up business use sqrt to make forecasts?: for Starting capital and inventory

Can you make a video explaining how to solve cubic equations using Ruffini method?

@carultch

6 ай бұрын

He has. Look for synthetic division and the rational roots theorem, in his playlists.

You should do a short about the most simplified equation for (x+1)^3 is actually 2.25(x^3)+2.25(x^2), because not a lot of people know that.

@carultch

6 ай бұрын

Not true at all. (x + 1)^3 is a standard cubic, translated to the left by 1 unit. It has a flat inflection point, that is a thrice-repeated root. 2.25*x^3 + 2.25*x^2 is a "roller coaster" cubic, with a negative slope at the inflection point. This cubic has a repeated root at the origin, and a distinct root at x=-1. Those aren't even close to being simplified forms of each other.

Pls Sir you could've show the solution too

What if we take the - sqrt of -x? Do we get it to flip both x and y axis, or do we end up cancelling each other out and going back to what we started with?

@mauriciocastillo7302

6 ай бұрын

In this case, we notice we need negative x values for this to be true. We also notice we multiplied the original function times negative one. Meaning, we flip across the x axis because of multiplying by -1, and we flip across the y axis as we need negative values for it to output real numbers.

Love your videos hate chalk

No teacher taught these equations like this back at school. Everything in maths is a lie 😭

During my school years I never had a teacher who taught. These were foreign to us. So pity hey 😂

When would it be graphed in the second quadrant?

@carultch

6 ай бұрын

Either when you translate it to the left, by adding to x inside the square root function, or when you reflect it to the left, by having -x inside the square root function. Technically, it does exist on the left side of the vertical axis, but its output is hidden in another dimension. See imaginary and complex numbers.

what reasoning is behind the rule that sqrt(x) is only +x ?

@carultch

6 ай бұрын

It's just a convention. Since it is most common that we are interested in the positive root, and the negative root is likely a non-physical root, we came up with the convention that radical signs imply the positive root if not otherwise specified. Negative roots aren't always non-physical, but it is most likely the case that the positive root is the only one of interest to an application of square roots. Hence, an interest in both roots, requires you to specify otherwise.

@nerd5865

5 ай бұрын

sqrt denotes to the principle root that's literally it

@ThomasHaberkorn

5 ай бұрын

that figures@@nerd5865

This guy is so up the math ladder I cant even see his ass. Shit am maybe in the 5 th rung imagine if everyone was like this guy I mean i think most problems would be solved No way one can a waste if one knows that much math

Are you real ? I mean are you really real ?

Third example would not work pretty sure 😂

@AizenSosukesama

6 ай бұрын

It would,whats wrong?I can explain

@TerminallyOnline92

6 ай бұрын

@@AizenSosukesama root of a negative number is imaginary result however we’re in the real plane so imaginary results do not exist and can not be shown on a graph like that. For another function other than root like trigonometric functions this would work but for roots it does not because of complex result

@juliuscaesar5098

6 ай бұрын

Nah it will work

@AizenSosukesama

6 ай бұрын

@@TerminallyOnline92 Thing is that in root(-x) here x can be a negative value,eg -5.Then it will be root(-(-5)) which is root(5) and we can show it on the real plane.Ofc for positive values of x it doesnt hve a real solution but that is why in the video the graph is only on negative x axis

@TerminallyOnline92

6 ай бұрын

@@AizenSosukesama oh yeah thx why didn’t I see that

Why didn't the last one have a green axis but got drawn by a chalk instead?

(1024)²⁰²⁴ is divided by 7 then what will be the reminder?? ??????? Can you please teach how to slove this type of questions

@lovejoy2438

6 ай бұрын

To find the remainder when (1024)^2024 is divided by 7, let's simplify the expression using modular arithmetic. First, notice that 1024 is congruent to 3 (mod 7) because 1024 ≡ 3 (mod 7). This means that 1024 and 3 leave the same remainder when divided by 7. Now, we can rewrite (1024)^2024 as (3)^2024 because 1024 ≡ 3 (mod 7). Next, we need to determine the remainder when (3)^2024 is divided by 7. To do this, we can use the property of modular exponentiation, which states that (a^b) mod c is congruent to [(a mod c)^b] mod c. Applying this property, we find that (3)^2024 mod 7 is congruent to [(3 mod 7)^2024] mod 7. Since 3 mod 7 equals 3, the expression simplifies to (3^2024) mod 7. Now, let's find the remainder: 3^1 ≡ 3 (mod 7) 3^2 ≡ 2 (mod 7) 3^3 ≡ 6 (mod 7) 3^4 ≡ 4 (mod 7) 3^5 ≡ 5 (mod 7) 3^6 ≡ 1 (mod 7) We notice that the remainders repeat every 6 powers of 3. Since 2024 is divisible by 6 (2024 ÷ 6 = 337 remainder 2), we can conclude that (3^2024) mod 7 is congruent to (3^2) mod 7. Therefore, (1024)^2024 mod 7 is congruent to 2. Thus, the remainder when (1024)^2024 is divided by 7 is 2.

@SwapnenduNayak

6 ай бұрын

@@lovejoy2438 thank you for your help, your explanation is very helpful to me