Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
Hi I really need help!
Date Posted: 2 years ago
Views: 189
Alternative approach, using discriminant of roots
What this question means is that for the values of y which can be "taken", a horizontal line y = k can be drawn to intersect the curve.
If the random horizontal line y = k does not intersect the curve, it means that the curve does not exist for that value of y.
But what values of y can we take? We are not sure. Let's allow the curve to intersect a random line y = k. Regardless of whether the line will cut the curve or not, we equate them together - to denote the fact that we are trying to put them on the same coordinate axes.
Then, we have
x² + 2x + 13 = k (x - 1)
x² + 2x + 13 = kx - k
x² + 2x - kx + 13 + k = 0
For some values of k, the horizontal line will cut the curve. For others, it won't be. Plotting a graph on graphing software confirms that the curve will not exist for some values of y.
With this in mind, the maximum number of times a horizontal line can intersect the curve will be two. So, we can model this using discriminant of roots.
If the intersection really happens (at one or two points),
Discriminant ≥ 0
(2 - k)² - 4 (1) (13 + k) ≥ 0
4 - 2k - 2k + k² - 52 - 4k ≥ 0
k² - 8k - 48 ≥ 0
(k + 4) (k - 12) ≥ 0
k ≤ -4 or k ≥ 12
So, any horizontal line y = k, where k ≤ -4 or k ≥ 12, will cut the curve at least once. These are the values of y that the curve can take, so
y ≤ -4 or y ≥ 12
What this question means is that for the values of y which can be "taken", a horizontal line y = k can be drawn to intersect the curve.
If the random horizontal line y = k does not intersect the curve, it means that the curve does not exist for that value of y.
But what values of y can we take? We are not sure. Let's allow the curve to intersect a random line y = k. Regardless of whether the line will cut the curve or not, we equate them together - to denote the fact that we are trying to put them on the same coordinate axes.
Then, we have
x² + 2x + 13 = k (x - 1)
x² + 2x + 13 = kx - k
x² + 2x - kx + 13 + k = 0
For some values of k, the horizontal line will cut the curve. For others, it won't be. Plotting a graph on graphing software confirms that the curve will not exist for some values of y.
With this in mind, the maximum number of times a horizontal line can intersect the curve will be two. So, we can model this using discriminant of roots.
If the intersection really happens (at one or two points),
Discriminant ≥ 0
(2 - k)² - 4 (1) (13 + k) ≥ 0
4 - 2k - 2k + k² - 52 - 4k ≥ 0
k² - 8k - 48 ≥ 0
(k + 4) (k - 12) ≥ 0
k ≤ -4 or k ≥ 12
So, any horizontal line y = k, where k ≤ -4 or k ≥ 12, will cut the curve at least once. These are the values of y that the curve can take, so
y ≤ -4 or y ≥ 12
Thank you for sharing!
See 1 Answer
The turning point, in itself, might not be sufficient to make the conclusion.
We somehow must be able to draw the conclusion that (5, 12) is a minimum point while (-3, -4) is a maximum point. We can rely on a graph to do this, or we just simply use regular brute force calculations to check for the nature of the turning point.
Added an alternative approach to the main question chat box.
This question looks more of a JC1 H2 Maths question than a Sec 3 A Maths question.
We somehow must be able to draw the conclusion that (5, 12) is a minimum point while (-3, -4) is a maximum point. We can rely on a graph to do this, or we just simply use regular brute force calculations to check for the nature of the turning point.
Added an alternative approach to the main question chat box.
This question looks more of a JC1 H2 Maths question than a Sec 3 A Maths question.
360 Tutoring Program