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why is only -9 substituted into the equation on 13b and not 1?
x = 1 is for the original point given in the question. We want the other point where the normal intersects the curve.
@@1stClassMaths Oh ok thank you! Yes that does make more sense
Hi sir. Today I learn something like (y-y1)=m(x-x1). I forgot when to use this, maybe you could explain
thanks
xdd
Can you use other differentiation techniques like quotient rule chain rule product rule etc in this exam?
I expect so yes.
I'm confused on question 13b, why do you substitute -9 into the original equation, not the equation of the normal?
You can substitute into either. Since both the curve and normal intersect here it doesn’t matter which one we choose.
I think you have got the last part of 13a wrong. It should be -4/3 not -2/3
Not sure you and I are looking at the same question 13?
I have already replied to you. I don't think you are looking at question 13a
why is only -9 substituted into the equation on 13b and not 1?
x = 1 is for the original point given in the question. We want the other point where the normal intersects the curve.
@@1stClassMaths Oh ok thank you! Yes that does make more sense
Hi sir. Today I learn something like (y-y1)=m(x-x1). I forgot when to use this, maybe you could explain
thanks
xdd
Can you use other differentiation techniques like quotient rule chain rule product rule etc in this exam?
I expect so yes.
I'm confused on question 13b, why do you substitute -9 into the original equation, not the equation of the normal?
You can substitute into either. Since both the curve and normal intersect here it doesn’t matter which one we choose.
I think you have got the last part of 13a wrong. It should be -4/3 not -2/3
Not sure you and I are looking at the same question 13?
I think you have got the last part of 13a wrong. It should be -4/3 not -2/3
I have already replied to you. I don't think you are looking at question 13a