## Question

secondary 3 | A Maths

##### Chelsia

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idk how to do

Date Posted: 2 months ago
Views: 18
Eric Nicholas K
2 months ago
Chelsia, it is noteworthy to know that ‘always positive’ graphs and ‘always negative’ graphs never ever cut the x-axis. For BOTH cases, b2 - 4ac < 0.

To distinguish the two cases, we look at the coefficient of x2. ‘Always positive’ graphs can only be exhibited by smiley face graphs, or a > 0. Similarly, ‘always negative’ graphs can only be exhibited by sad face graphs, or a < 0.

Whenever you see a question like ‘find the range of values of k (note: not x) for which the curve is always positive for all real values of x’, this means the graph will never cut the x-axis at all, so you must immediately identify that b2 - 4ac < 0 (and a > 0).

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Hi Chelsia! Here are my workings for this question. Let me know if my workings are unclear and I will do my best to explain them again.
Date Posted: 2 months ago
Chelsia
2 months ago
I dont understand the steps
Eric Nicholas K
2 months ago
When a curve is always negative, it means that the curve is always below the x-axis. Only sad face graphs are capable of being fully below the x-axis. Smiley face graphs can never be ‘fully below the x-axis’.

Being fully below the x-axis means that the curve will not ever cut the x-axis. In turn, this means that there are no solutions to the equation y = 0. No solutions means no real roots; hence, b2 - 4ac must be < 0.

You will need to find out what a, b and c are from the equation. Here, a = k + 1, b = -2k and c = 3k. From there, you will need to form an inequality and solve for the inequality. Note that multiplying or dividing both sides of an inequality by a negative number changes the sign of the inequality (you will need to have the necessary knowledge from E Maths first).

You will need to bear in mind that only sad face graphs can be 'always negative'. We look at the coefficient of x2 to identify whether a graph is smiley face or sad face. For sad face graphs, the coefficient must be a negative number. In other words, a < 0, or in this case, k + 1 < 0. Where necessary (it does happen in this question eventually), we will need to reject some values of k.