how come the coordinates arent (5,0) for the third question? doesn't it become positive either way when it's a modulus function? or is it because we are completing the square? thanks
If you substitute x=5 into the equation y = |2x + 10| you get y = |2*5+10| = 20, not zero. When you substitute a negative number in, you don't change it to being positive before you substitute it in.
![Kumamoto Masters Japan 2024 | Gregoria Mariska Tunjung (INA) [5] vs. Akane Yamaguchi (JPN) [2] | F](http://i.ytimg.com/vi/EIjZybhsWqA/mqdefault.jpg)
comparing it to completing the square - such a good method, never heard of it before. Thanks!
this guy is a legend
I love your videos my dawg. Praying they get me through my exam today. Big love
Nicely explained in easy way
Cheers mate very good explanation love it
Nicely edited
why is he so damn underrated
Good job..keep it up.. well wishes from Pakistan❤
Wow! Very clearly explained! :)
Nice video, thanks❤
Good sir
Why isn't the vertex (5,0) for Q3? I though you have to take the modulus (positive) of 5
But x=-5 makes what is inside the modulus zero, and hence y is zero at that point. So the vertex is at (-5,0).
how come the coordinates arent (5,0) for the third question? doesn't it become positive either way when it's a modulus function? or is it because we are completing the square? thanks
If you substitute x=5 into the equation y = |2x + 10| you get y = |2*5+10| = 20, not zero. When you substitute a negative number in, you don't change it to being positive before you substitute it in.
How come it hasn’t been reflected for when y
When x = 3, you get y = |3 - 3| - 3 = -3, so the graph definitely goes below the x-axis?