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#1
How do you know to subtract 48 from ii

I know that this is the area of the 'box'that the curve is enclosed in but how do you know if the curve is below the x axis so the length of the y axis may be longer than 12

Question 5iii
http://pmt.physicsandmathstutor.com/...oints%20QP.pdf

Mark scheme question 5iii:http://pmt.physicsandmathstutor.com/...oints%20MS.pdf

Why also does it work only when I substitute x into the factorised form to get y but not into the original expanded form to get y? I get -124 when I substitute it into the original equation

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2 years ago
#2
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#3
(Original post by Justvisited)
Try now
http://pmt.physicsandmathstutor.com/...oints%20QP.pdf
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2 years ago
#4
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2 years ago
#5
(Original post by esmeralda123)
How do you know to subtract 48 from ii

I know that this is the area of the 'box'that the curve is enclosed in but how do you know if the curve is below the x axis so the length of the y axis may be longer than 12

Question 5iii
http://pmt.physicsandmathstutor.com/...oints%20QP.pdf

Mark scheme question 5iii:http://pmt.physicsandmathstutor.com/...oints%20MS.pdf

Why also does it work only when I substitute x into the factorised form to get y but not into the original expanded form to get y? I get -124 when I substitute it into the original equation

How you know, is by sketching it, first considering that it's a negative quadratic so a "frowny face". You can factorise it to confirm that its roots are at x = -1 and x = 7, so obviously between these x values the curve is above the x axis. As it happens, it'll also be above the line y = 12 between x = 1 and x = 5, so they're asking for this "rocket cone" shaped area, which will be the area under the curve minus the rectangle bounded by y = 0, y = 12, x = 1 and x = 5, with length 4 and height 12, hence area 48.

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