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secondary 2 | Maths
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how do I find the answer for part c? tysm
Date Posted: 3 years ago
Views: 176
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done
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Pls ignore right side if you havent learnt differentiation
Date Posted:
3 years ago
For the Sec 2 level the “take average of x-intercepts” approach should suffice. Differentiation is probably a little too advanced for the student.
The curve is symmetrical. The minimum point lies on the line of symmetry.
Both x-intercepts (which are mirror images of each other) are equidistant from the line of symmetry. So the midpoint between them lies on this line as well.
x-coordinate of the midpoint = (3 + 7)/2 = 10/2 = 5
This tells you that the line of symmetry has equation x = 5. Which means that the minimum point has an x-coordinate of 5 .
When x = 5, y = (5 - 3)(5 - 7) = (2)(-2) = -4
So the minimum point has coordinates (5,-4)
Both x-intercepts (which are mirror images of each other) are equidistant from the line of symmetry. So the midpoint between them lies on this line as well.
x-coordinate of the midpoint = (3 + 7)/2 = 10/2 = 5
This tells you that the line of symmetry has equation x = 5. Which means that the minimum point has an x-coordinate of 5 .
When x = 5, y = (5 - 3)(5 - 7) = (2)(-2) = -4
So the minimum point has coordinates (5,-4)
done
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find min or maximum point, dy/dx curve at let it = 0. hope it helps!
Date Posted:
3 years ago
“dy/dx” might be slightly too advanced for a Secondary 2 student.
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