• Making the Shifts
• Task
• If we put the equation $4x-y=c$ in the form $y=4x-c$, we see that the graph has slope $4$. The slope of the graph of $y=2x+d$ is $2$. So the steeper line, $l$, is the one with equation $y=4x-c$, and therefore $-c$ is the $y$-coordinate of the point where $l$ intersects the $y$-axis. The other line, $m$, is the one with equation $y=2x+d$, so $d$ is the $y$-coordinate of the point where $m$ intersects the $y$-axis.

  Commentary:

  This task requires students to use the fact that on the graph of the linear equation $y=ax+c$, the $y$-coordinate increases by $a$ when $x$ increases by one. Specific values for $c$ and $d$ were left out intentionally to encourage students to use the above fact as opposed to computing the point of intersection $(p, q)$, and then computing respective function values to answer the question.

  Solution:

  1. If we put the equation $4x-y=c$ in the form $y=4x-c$, we see that the graph has slope $4$. The slope of the graph of $y=2x+d$ is $2$. So the steeper line, $l$, is the one with equation $y=4x-c$, and therefore $-c$ is the $y$-coordinate of the point where $l$ intersects the $y$-axis. The other line, $m$, is the one with equation $y=2x+d$, so $d$ is the $y$-coordinate of the point where $m$ intersects the $y$-axis.
  2. The line $l$ has slope $4$. So on $l$, each increase of one unit in the $x$-value produces an increase of $4$ units in the $y$-value. Thus an increase of $2$ units in the $x$-value produce an increase of $2⋅4=8$ units in the $y$-value. The line $m$ has slope $2$. So on $L_2$, each increase of $1$ unit in the $x$-value produces an increase of $2$ units in the $y$-value. Thus an increase of $2$ units in the $x$-value produces an increase of $2⋅2=4$ units in the $y$-value. Thus your $y$-value would be $8-4=4$ units larger than my $y$-value.
• Additional Thoughts

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