0

Why is the graph of $x-1>y$ like this? I cannot get it intuitively. Can someone help? Thanks.

enter image description here

ACB
  • 3,713
Aleph
  • 425
  • 2
  • 10
  • I meant to ask that why all points below that half plane give $x-1>y$? – Aleph Feb 04 '22 at 18:33
  • Perhaps look at the equivalent inequality $y<x-1$. – David Mitra Feb 04 '22 at 18:34
  • Check some? So, $(4,-1)$ gives $x-1 = 3$ which is indeed greater than $-1$. – Randall Feb 04 '22 at 18:34
  • @Randall how does checking some will verify us that all points satisfy that inequality? – Aleph Feb 04 '22 at 18:35
  • @Aqua it doesn't, but it builds your intuition on why this is correct – Randall Feb 04 '22 at 18:35
  • @Randall Oh ok. Is there a method to ensure that all points below that half plane will satisfy the inequality? – Aleph Feb 04 '22 at 18:37
  • 1
    $x$ approaching $+\infty$ is to the right, while $y$ approaching $+\infty$ is upward. – user2661923 Feb 04 '22 at 18:41
  • 2
    Look at the vertical line through $x=a$. The points satisfying $y<x-1$ with $x=a$ are those on the vertical line and below the line $y=x-1$. – David Mitra Feb 04 '22 at 18:43
  • 1
    Are you comfortable with coming up with a visual representation of $x-1>y$ for a specific value of $y$? For example, coming up with a visual representation of $x-1>0$? Coming up with a visual representation of $x-1>1$, for $x-1>2$ and so on? Now... think of these as horizontal slices of your final graph and stack them up onto one another and generalize to arbitrary $y$ as opposed to specific individual values of $y$... – JMoravitz Feb 04 '22 at 18:45
  • I am still confused how you are going about it.. Can you please explain in an easier way? – Aleph Feb 04 '22 at 18:58
  • I asked several leading questions. Focus on the second for now... Are you or are you not comfortable coming up with a visual representation of $x-1>0$? – JMoravitz Feb 04 '22 at 19:01
  • We have $y<x-1$. First draw the graph of $y=x-1$, it is a line contains all the points $(x,y)$ so that value of $y$ on each point is equal to $x-1$. Now, value of $y$ in any point above the line is greater than value of $y$ on the line (which is equal to $x-1$) and vise versa. So $y>x-1$ denotes all the point above the line $y=x-1$ and $y<x-1$ denotes all the points below that line. – Etemon Feb 04 '22 at 20:12

1 Answers1

1

Think in terms of horizontal slices and specific values of $y$. When $y$ is specifically equal to $0$ the inequality becomes $x-1>0$ or rearranged, $x>1$. So, think of graphing $x>1$ on the number line.

Do the same for when $y=1$ this becomes $x-1>1$ or rearranged $x>2$.

Imagine doing this for "many" such values of $y$ and coming up with "many" stacked inequalities on the numberline.

stacked inequalities

Now, make sure that each horizontal slice is at the height corresponding to it's respective $y$-value

JMoravitz
  • 79,518