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junior college 1 | H2 Maths
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Good evening!! Can someone pls help me with this? I don’t understand how to go about doing this when it’s all a,b,c,d :-( Thank you so much!
Date Posted: 4 years ago
Views: 241
Good evening Sonia! I will have a look at this question later if no one else has done them yet.
Good evening Sonia!
Since we are asked to do differentiation for part (i), we proceed by differentiating the expression via the quotient rule.
Sonia, do you recall how to do the chain, product and quotient rule? These concepts are necessary in the second half of the year when you learn differentiation and integration in greater detail. If not, fret not; I have included my workings below for the differentiation.
After simplifying our expression, we still need to explain why the graph has no turning points. We know that at any turning point, dy/dx = 0. The fact that the graph has turning points means that there is no value of x in the graph for which dy/dx = 0. We simply explain, using our knowledge of perfect squares etc, why we are unable to equate dy/dx to zero no matter what.
For part (ii), we simply follow our result in part (i), except this time with ad - bc = 0, and make some conclusion there. As it turns out, our expression for dy/dx ends in a fraction with the numerator being ad - bc and the denominator being (cx + d)^2. Of course, the numerator is going to be zero regardless of the value of x because there is simply no x in the numerator. We will realise that the only graph which contains dy/dx = 0 at ALL points of the graph is a horizontal line.
For part (iii), we simply put in the numbers for a, b, c and d and make further conclusions.
For part d, we attempt to obtain the intercepts and the asymptotes. The intercepts are easy, since we simply set x = 0 for the y-intercept and y = 0 for the x-intercept. You can follow my workings to see how I write out my asymptotes. I am sure you have questions regarding the asymptoted (yes, because I know I rushed my workings there).
Sonia, let me know if you need more explanation.
Since we are asked to do differentiation for part (i), we proceed by differentiating the expression via the quotient rule.
Sonia, do you recall how to do the chain, product and quotient rule? These concepts are necessary in the second half of the year when you learn differentiation and integration in greater detail. If not, fret not; I have included my workings below for the differentiation.
After simplifying our expression, we still need to explain why the graph has no turning points. We know that at any turning point, dy/dx = 0. The fact that the graph has turning points means that there is no value of x in the graph for which dy/dx = 0. We simply explain, using our knowledge of perfect squares etc, why we are unable to equate dy/dx to zero no matter what.
For part (ii), we simply follow our result in part (i), except this time with ad - bc = 0, and make some conclusion there. As it turns out, our expression for dy/dx ends in a fraction with the numerator being ad - bc and the denominator being (cx + d)^2. Of course, the numerator is going to be zero regardless of the value of x because there is simply no x in the numerator. We will realise that the only graph which contains dy/dx = 0 at ALL points of the graph is a horizontal line.
For part (iii), we simply put in the numbers for a, b, c and d and make further conclusions.
For part d, we attempt to obtain the intercepts and the asymptotes. The intercepts are easy, since we simply set x = 0 for the y-intercept and y = 0 for the x-intercept. You can follow my workings to see how I write out my asymptotes. I am sure you have questions regarding the asymptoted (yes, because I know I rushed my workings there).
Sonia, let me know if you need more explanation.
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Good evening Sonia! Here are my workings for this question.
Date Posted:
4 years ago
Hi Eric! Thanks so much for ur detailed explanations but may i ask how u simplified the y function until it became 3/2 + ... ?
I don’t quite understand how u did that because normally i would do long division, however, i can’t seem to do long division for this qns..
I don’t quite understand how u did that because normally i would do long division, however, i can’t seem to do long division for this qns..
Good morning Sonia!
Yes, we do long division on the (3x - 7) / (2x + 1) expression as we would normally. The quotient is 3/2. The remainder, if you have done correctly, is -7 - 3/2 * 1 = -17/2. You still remember how to do long division from last year correct?
So basically we can do long division on the numbered expressions. Doing on the lettered one in the original question is possible, but the expression gets complicated.
After years of practising with long division and understanding the operations, I tend to be able to divide the expression the way I did in the question (without actually writing out the full division).
Sonia, I’m so sorry for making you confused with my workings!
Yes, we do long division on the (3x - 7) / (2x + 1) expression as we would normally. The quotient is 3/2. The remainder, if you have done correctly, is -7 - 3/2 * 1 = -17/2. You still remember how to do long division from last year correct?
So basically we can do long division on the numbered expressions. Doing on the lettered one in the original question is possible, but the expression gets complicated.
After years of practising with long division and understanding the operations, I tend to be able to divide the expression the way I did in the question (without actually writing out the full division).
Sonia, I’m so sorry for making you confused with my workings!
No it’s totally fine! There’s no need to apologize, Thank you once again :-)
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