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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 346
The other two tutors have explained this question using a Sec 4 A Maths approach of differentiation. Let me attempt to explain this using a Sec 3 A Maths approach of completing the square.
Let y = px2 - 6x + q.
Now, obviously p cannot be zero or this will not be quadratic. If p is positive, the graph will be “smiley face” and will have a minimum point ie minimum value. If p is negative, the graph will be “sad face” and will have a maximum point ie maximum value.
Though p and q can take several values, a specific value of p will lead to a specific value of q.
In the simplest case, let p = 1. This leads to a minimum value.
y = x2 - 6x + q
= x2 - 6x + (-3)^2 - (-3)^2 + q
= (x - 3)^2 - 9 + q
= (x - 3)^2 + q - 9
The expression outside the perfect square brackets ie “q - 9” will result in the maximum or minimum value of the graph (here, a minimum since p = 1).
For the minimum value to equal 8,
q - 9 = 8
q = 17
So a pair of possible values for p and q to achieve a minimum of 8 is p = 1, q = 17.
All we need to do is to change p to -1, but keep q as 17, for the minimum to change into maximum at y = 8.
So a pair of possible values for p and q to achieve a maximum of 8 is p = -1, q = 17.
Other possible values, using similar approaches, are
(Min) p = 3, q = 11
(Max) p = -3, q = 11
(Min) p = 2, q = 12.5
(Max) p = -2, q = 12.5
And many more, including decimal values of p.
Let y = px2 - 6x + q.
Now, obviously p cannot be zero or this will not be quadratic. If p is positive, the graph will be “smiley face” and will have a minimum point ie minimum value. If p is negative, the graph will be “sad face” and will have a maximum point ie maximum value.
Though p and q can take several values, a specific value of p will lead to a specific value of q.
In the simplest case, let p = 1. This leads to a minimum value.
y = x2 - 6x + q
= x2 - 6x + (-3)^2 - (-3)^2 + q
= (x - 3)^2 - 9 + q
= (x - 3)^2 + q - 9
The expression outside the perfect square brackets ie “q - 9” will result in the maximum or minimum value of the graph (here, a minimum since p = 1).
For the minimum value to equal 8,
q - 9 = 8
q = 17
So a pair of possible values for p and q to achieve a minimum of 8 is p = 1, q = 17.
All we need to do is to change p to -1, but keep q as 17, for the minimum to change into maximum at y = 8.
So a pair of possible values for p and q to achieve a maximum of 8 is p = -1, q = 17.
Other possible values, using similar approaches, are
(Min) p = 3, q = 11
(Max) p = -3, q = 11
(Min) p = 2, q = 12.5
(Max) p = -2, q = 12.5
And many more, including decimal values of p.
See 3 Answers
done
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There will be a range of answers but integer numbers look nicer haha
Date Posted:
4 years ago
done
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No need differentiation. Hope this is the one you were looking for.
Date Posted:
4 years ago
Let y = px^2 - 6x + q
Im assuming you know differentiation here.
dy/dx = 2px - 6.
To find the min/max point, we let dy/dx = 0
So, we get that x = 3/p.
Now, we substitute that into y, and we get
y = 9/p - 18/p + q
y = -9/p + q
We want to find sets of values of p and q such that y = 8 is the min or max point.
So, we let y = 8.
So, 8 = -9/p + q
q = 8 + 9/p
Suppose we let p = 1 , q = 17. This is for the lowest point where y = 8.
Suppose now, we let p = -1, q = -1. This is for the highest point where y = 8.
Recall that if the coefficient of x^2 is positive, the curve looks like a smiley face, hence, it will have a lowest point.
Similarly, when the coefficient of x^2 is negative, the curve looks like a sad face, hence it will have a highest point.
Im assuming you know differentiation here.
dy/dx = 2px - 6.
To find the min/max point, we let dy/dx = 0
So, we get that x = 3/p.
Now, we substitute that into y, and we get
y = 9/p - 18/p + q
y = -9/p + q
We want to find sets of values of p and q such that y = 8 is the min or max point.
So, we let y = 8.
So, 8 = -9/p + q
q = 8 + 9/p
Suppose we let p = 1 , q = 17. This is for the lowest point where y = 8.
Suppose now, we let p = -1, q = -1. This is for the highest point where y = 8.
Recall that if the coefficient of x^2 is positive, the curve looks like a smiley face, hence, it will have a lowest point.
Similarly, when the coefficient of x^2 is negative, the curve looks like a sad face, hence it will have a highest point.
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