it’s completing the square when the question asks for the turning point of a graph, or when it says give the answer in the form (x+a)^2 +b. it’s factorising if the question says solve. or work out the x intercepts, or work out the solutions to a graph
@@1stClassMaths how about the same equation being equal to zero, i.e. f'(x)= 128x^2 +125x+1000=0 for x>0 where f'(x) is dy/dx of another equation.Then what should we do?
@@1stClassMaths At first, I thought it was wrong because it's for decreasing and I thought that the values below 0 had both decreasing and increasing, since it decreases to a minimum point, but then rises up. In short, I thought the decreasing part would be the left hand side of the graph, not the part below 0
Actually, for a decreasing function, it's where the gradient of the graph is negative or less than zero. It's not where it is below the x-axis. Below the x-axis tells where the graph is negative and not where the gradient is negative. Actually, the gradient could be negative above the x-axis as well.
Thanks man.
Cheers
How do you know when to factorise and when to complete the square ?
it’s completing the square when the question asks for the turning point of a graph, or when it says give the answer in the form (x+a)^2 +b. it’s factorising if the question says solve. or work out the x intercepts, or work out the solutions to a graph
Also, just to add, if you cant factorise with factors then you can resort to completing the square or other methods
what if the dy/dx is a quadratic eq and the its equal to zero? E.g 128x^2 +125x+1000 for x>0. Is it increasing/decreasing/neither ?
128x^2 +125x+1000 is > 0 for all values of x so the function is always increasing.
@@1stClassMaths how about the same equation being equal to zero, i.e. f'(x)= 128x^2 +125x+1000=0 for x>0 where f'(x) is dy/dx of another equation.Then what should we do?
That function cannot be equal to 0 for any x value. It is always positive.
@@1stClassMaths thanks
Why cant it be 0 is greater than "and equal to" x..etc.. Is it bcuz of dy/dx < 0?
what if you have to work out the values of x for the function y= 2x-3 which is increasing?
All values. dy/dx = 2 > 0 (Its also a straight line graph so if you think about the graph its always increasing)
@@1stClassMaths makes sense, thank you
I believe it is incorrect from 4:39 onwards for the final example
Im not sure though
I don’t think so it looks ok to me. What do you think it wrong?
@@1stClassMaths At first, I thought it was wrong because it's for decreasing and I thought that the values below 0 had both decreasing and increasing, since it decreases to a minimum point, but then rises up. In short, I thought the decreasing part would be the left hand side of the graph, not the part below 0
Ah ok. Remember it isn’t when this graph is decreasing but the graph of the original function. If you graph the cubic it doe decrease from 0 to 2.
Actually, for a decreasing function, it's where the gradient of the graph is negative or less than zero. It's not where it is below the x-axis. Below the x-axis tells where the graph is negative and not where the gradient is negative. Actually, the gradient could be negative above the x-axis as well.
for q1, would 0 < x < 2/3 still be correct or only x
Only < 2/3 since negative values are still ok.
what is a critical value?
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