mate, this video has helped me so much I just hate how my maths teacher mentions things but doesn't teach them and then formally tests class on things we haven't been taught. ugh. you have given me a glimpse at passing this test, thankyou @Jeremy Klassen
Thanks sooooo much omg. I was so stressed I had no idea what to do in that situation and now I finally found the video and I’m ready for my test tmrw!!!!!! Thankssss!!!!
dividing a negative by a negative is the same as dividing a positive by a positive. -5/-2 is the same as 5/2 so we just remove the negatives to make things easier. If you wrote -5/-2 you'd still be correct.
If you were to have say *4=x+3* for the part at 2:33 would you find the square root of *4* (making 2) and then at the add and subtract sign or just leave it as *4*?
guys what if the value u bring over becomes a negative cuz then u cant square root a negative... right? -5 = (x+2)² (square root) -5 = x+2 u can't square root a negative
To find x intercepts we substitute a 0 for the y variable. This gives us 0=2(x-3)(x+1). On the right side of this equation are three factors. If their product is 0, one of these is 0. It can't be the 2, so it must be either x-3 or x+1. Turns out it could be either one. If x=3, then x-3 becomes zero. If x=-1, then x+1 becomes zero. So these are the x intercepts: -1, 3. To get the vertex we could approach this in a few different ways. I think the most obvious is to realize that the vertex will be located on the parabola on the axis of symmetry and the axis of symmetry cuts the parabola in half. This means that it must be half way between -1 and 3. If you find the average of -1 and 3 you get x=1. So 1 is the x coordinate of the vertex. To get the y coordinate we use the original equation of the parabola. So y=2(1-3)(1+1)=-8. So now we know the vertex must be the point (1,-8).
You have the two x coordinates, 4+the square root of 5/2 and 4- the square root of 5/2. You were also given, at the beginning of the problem, a function for y in terms of x. You take these values of x and substitute them in to find the corresponding values of y. Now, that being said, because we are finding the x-intercepts the value of y should be zero in both cases. So our points would be (4+the square root of 5/2, 0) and (4-the square root of 5/2,0).
@@raymondbao1982 ok. That sort of depends on what your teacher is expecting. Some don't mind if you leave the 5/2 under the radical, but some will want you to rationalize the denominator. You might have to multiply numerator and denominator by root 2 to get 1/2(root 10). I am not sure if this helps at all. I might have to make another video. Let me know
+yasser Hussein If the 5 in the original function had been negative, then when it was brought to the other side of the equation we would have made it positive. Then we would have divided by -2 and the result would have been negative. You are correct that this leaves us with a problem, because there are no real solutions to the square root of a negative. This is not, however a problem, because if the 5 was negative in the original, then the vertex would have been (-3, -5) and the parabola would have opened in the negative direction - thus it wouldn't have had x-intercepts anyways. I hope that helps.
In my class they don’t give x or y so how do I most problems look like this - h(x) = 1x^2 -50x + 60 Just an example problem but I’m in online school so I can’t ask anyone :/
Well in this case, with it expressed in that form, to find the x-intercepts I would let h(x) be zero and I would use the quadratic formula. Does that make sense?
The x-intercepts are the points where a curve crosses the x-axis. Another way of thinking about them is that they are the values of the independent variable, when the dependent variable takes on a value of 0. ie it is the value of x when y is 0. I hope that helps.
Now I understand your question. The x intercepts would be -3+(5/2)^(1/2) and -3-(5/2)^(1/2), or if you wanted a decimal approximation -1.42 and -4.58. The addition sign over the subtraction sign implies two solutions can be found by first adding the two number and then subtracting the two numbers.
The equation is in vertex form so we can see immediately that the vertex is (-3,5) if I am remembering it correctly and then because of the negative two in front it opens down with a vertical stretch by a factor of two.
Oh. Okay. Well you're right. The x-intercepts aren't going to be very useful to us in that way. Well we know the vertex is (-3,5). What you could do, if you need some points to help you plot the parabola is pick a couple of x-coordinates that are symmetrically spaced away from the vertex, say x=-5 and -2. Substitute these into the original quadratic to find their corresponding y-coordinates and plot those points. When you do and then connect the vertex through these points, you should get a fairly accurate representation of the shape of the parabola. I hope that makes sense.
Thank you. Your 3 minute video taught me more than my teacher did in the whole unit
literally watching this during my test lmao
Online test rocks!
Samee
@@enochsuperstar572 I'm sure a lot of people loved online classes just because they could afk. I've seen it before. It's their grade, not mine.
Thanks for the help, this really helped me with the math my teacher could barely explain.
mate, this video has helped me so much I just hate how my maths teacher mentions things but doesn't teach them and then formally tests class on things we haven't been taught. ugh. you have given me a glimpse at passing this test, thankyou @Jeremy Klassen
Almost 10 yrs later and your video helped me with my math assignment thank you!!!
thank you for explaining this so well! i was having problems until i came across your video, thanks so much :)
why is there only one x intercept , shouldn't there be 2 ?
Not sure if you still need this, but since it’s plus or minus, there’s two possible solutions in this answer
Dude thank you so much ! You legit saved my life ! Much love ~
Thanks sooooo much omg. I was so stressed I had no idea what to do in that situation and now I finally found the video and I’m ready for my test tmrw!!!!!! Thankssss!!!!
my man, my man
thanks for making this so simple and so short! big help!
God bless your soul for making this so simple ❤❤❤❤❤❤❤❤
Goooooood🌸🌹🌺🌹🌸🌸🌸🌸🌺
one question, why isnt it 5 over -2? im really confused about that part. Are you suppose to include the negative sign in front of the 2?
-5/-2 the negatives cancel out to give 5/2.
@@zaramahajan8603 actually it's bcs the number is going from × to ÷, not plus or minus so no change in number. Awfully due reply tho 😆😆
dividing a negative by a negative is the same as dividing a positive by a positive. -5/-2 is the same as 5/2 so we just remove the negatives to make things easier. If you wrote -5/-2 you'd still be correct.
🐐
Saved my ass
look at the top right, he got an A+
Awesome video! Thanks so much!
WOW. Thank you Thank you!
If you were to have say *4=x+3* for the part at 2:33 would you find the square root of *4* (making 2) and then at the add and subtract sign or just leave it as *4*?
guys what if the value u bring over becomes a negative cuz then u cant square root a negative... right?
-5 = (x+2)²
(square root) -5 = x+2
u can't square root a negative
Then there are no real roots, so no solutions depending what level of mathematics you are in
Thank you
You're the real MVP
what a savior
Thanks, I appreciate it.
How do I find the x intercepts and the vertex with y=2(x-3)(x+1)
To find x intercepts we substitute a 0 for the y variable. This gives us 0=2(x-3)(x+1). On the right side of this equation are three factors. If their product is 0, one of these is 0. It can't be the 2, so it must be either x-3 or x+1. Turns out it could be either one. If x=3, then x-3 becomes zero. If x=-1, then x+1 becomes zero. So these are the x intercepts: -1, 3. To get the vertex we could approach this in a few different ways. I think the most obvious is to realize that the vertex will be located on the parabola on the axis of symmetry and the axis of symmetry cuts the parabola in half. This means that it must be half way between -1 and 3. If you find the average of -1 and 3 you get x=1. So 1 is the x coordinate of the vertex. To get the y coordinate we use the original equation of the parabola. So y=2(1-3)(1+1)=-8. So now we know the vertex must be the point (1,-8).
Jeremy Klassen damn much appreciated for the response but I just finished the test and I didn’t see but hopefully I got part marks
Jeremy Klassen do you have a email or some contact if I have questions you could answer them for me ty very much
oh never mind the negatives cancel... sorry XD
Thank you so much
Watching this 10 minutes before my test💀
Thank you random internet guy
thank you!
how do you turn 4 ± the square root of 5/2=x into coordinate points?
You have the two x coordinates, 4+the square root of 5/2 and 4- the square root of 5/2. You were also given, at the beginning of the problem, a function for y in terms of x. You take these values of x and substitute them in to find the corresponding values of y. Now, that being said, because we are finding the x-intercepts the value of y should be zero in both cases. So our points would be (4+the square root of 5/2, 0) and (4-the square root of 5/2,0).
@@JeremyKlassenThePiMan how should i write the square root of 5/2 as a fraction
@@JeremyKlassenThePiMan i understand that i just am confused on how to write 5/2 square root as a fraction
@@raymondbao1982 ok. That sort of depends on what your teacher is expecting. Some don't mind if you leave the 5/2 under the radical, but some will want you to rationalize the denominator. You might have to multiply numerator and denominator by root 2 to get 1/2(root 10). I am not sure if this helps at all. I might have to make another video. Let me know
@@JeremyKlassenThePiMan she says to round it to the nearest tenth
THANK YOU!!!!
thank youuu so muchhh
Big thanks
Mans saved me
thanks, helped alot!
Its asking me to state the exact value of the coordinate, so is that okay?
thx really helped
Ik im a few years late but my teacher gave the two zeros and I trued using this and I didn’t get both of them
i love you
Nice thanks
Give me the F! Right the damn now!
Silent Key please let me know if there is something confusing or where it stops making sense. I'd really like to help.
Omg, ily
but what if u get a positive number over a negative number? what would happen then?
+yasser Hussein If the 5 in the original function had been negative, then when it was brought to the other side of the equation we would have made it positive. Then we would have divided by -2 and the result would have been negative. You are correct that this leaves us with a problem, because there are no real solutions to the square root of a negative. This is not, however a problem, because if the 5 was negative in the original, then the vertex would have been (-3, -5) and the parabola would have opened in the negative direction - thus it wouldn't have had x-intercepts anyways. I hope that helps.
In my class they don’t give x or y so how do I most problems look like this - h(x) = 1x^2 -50x + 60
Just an example problem but I’m in online school so I can’t ask anyone :/
Well in this case, with it expressed in that form, to find the x-intercepts I would let h(x) be zero and I would use the quadratic formula. Does that make sense?
Yes, thanks.
So what are the x intercepts?
The x-intercepts are the points where a curve crosses the x-axis. Another way of thinking about them is that they are the values of the independent variable, when the dependent variable takes on a value of 0. ie it is the value of x when y is 0. I hope that helps.
So would -3 and 5/2 be the x intercepts you used in the example (2:54)
Now I understand your question. The x intercepts would be -3+(5/2)^(1/2) and -3-(5/2)^(1/2), or if you wanted a decimal approximation -1.42 and -4.58. The addition sign over the subtraction sign implies two solutions can be found by first adding the two number and then subtracting the two numbers.
Is the y the same thing as h(x)? none of the ones Im doing have a y in them from the beginning
Yes.
@@JeremyKlassenThePiMan Thank you!
how do we graph that tho
The equation is in vertex form so we can see immediately that the vertex is (-3,5) if I am remembering it correctly and then because of the negative two in front it opens down with a vertical stretch by a factor of two.
What do I do when I need to graph these aren’t x intercept points it’s a radical I’m so confused
I'm not sure I am following you. Could you show me the question? Perhaps I could walk it through with you.
Jeremy Klassen He is asking for specific points to graph. -3 + 5/2 isn't quite exact as coordinate points, and he's confused on that.
Oh. Okay. Well you're right. The x-intercepts aren't going to be very useful to us in that way. Well we know the vertex is (-3,5). What you could do, if you need some points to help you plot the parabola is pick a couple of x-coordinates that are symmetrically spaced away from the vertex, say x=-5 and -2. Substitute these into the original quadratic to find their corresponding y-coordinates and plot those points. When you do and then connect the vertex through these points, you should get a fairly accurate representation of the shape of the parabola. I hope that makes sense.
I love you ...
are zeros and x-intercepts the same?
Yes. They come from two different ways of discussing what amounts to the dame thing
@@JeremyKlassenThePiMan Okay, thanks
r u laying down lol
Wasted my time
u just dumb fortnite kid