ATAR Notes: Forum
VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: elisekatherine on May 29, 2011, 01:53:19 pm
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I'm doing a general test tomorrow and need help with a few questions. The test is on graphing (linear) and is pretty simple, because it's open book. I'm revising now, and I'm having trouble with two questions.
1) Find the equation of a line parallel to y=-2x+1 and passing through the point (2,4)
2) Sketch the graph of y=2x and y=x/2 on the same set of axes. Show the line y=x also.
If anyone could help me with those that would be great. Thankyou :)
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1) well if the new line is parallel to y=-2x+1 then the gradient is the same. Remember the gradient is m in y=mx+c, so m=-2
so the new line is y=-2x+c. To find the c value sub in your point (2,4) and solve for c:
4=-2(2)+c so c=8
therefore the line is y=-2x+8
2) well y=x is a straight line through the origin with gradient 1.
y=2x is the same, but has gradient 2, so its steeper
y=x/2 is the same again but has a flatter gradient than y=x
in the picture, red is y=x, blue is y=x/2 and green is y=2x
remember to label the axis and you could label a point on each graph if you wanted, eg (1,1) for y=x
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1) well if the new line is parallel to y=-2x+1 then the gradient is the same. Remember the gradient is m in y=mx+c, so m=-2
so the new line is y=-2x+c. To find the c value sub in your point (2,4) and solve for c:
4=-2(2)+c so c=8
therefore the line is y=-2x+8
2) well y=x is a straight line through the origin with gradient 1.
y=2x is the same, but has gradient 2, so its steeper
y=x/2 is the same again but has a flatter gradient than y=x
in the picture, red is y=x, blue is y=x/2 and green is y=2x
remember to label the axis and you could label a point on each graph if you wanted, eg (1,1) for y=x
Thankyou very very much! I understand it perfectly now! (: