Old hsc question

AlonzoN

Junior Member
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Jul 31, 2021
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Find the values of k for which x^2 - 3x + (4 - 2k) is always positive.
 
You need to show us what work you've been able to do on this. We can't really help you that well unless we know. I can say this:
Hint: Look at the discriminant. What does that tell you about the nature of the roots?

-Dan
 
Have another look at my response to a previous post of yours which involved finding when a cubic function was increasing.
Same concept applies here.
 
My first try,

x^2 - 3x + (4-2k), if x =1

1^2 - 3*1 + 4 - 2k

2 - 2k

So we need it to be increasing,

2 - 2k <= 0

-2k <= -2

k >= 1,

but then I thought again that maybe

4 - 2k <= 0

-2k <= -4

k >= 2

I know I am doing SOMETHING horribly wrong.
 
Yes! Have you looked at my post on your other question as I suggested. Yo had commented "Thanks Heaps" so I assume you understood what I did.
 
my bad, other q, I think it would be 0 and 3 if it was just x^2 and -3x, but I know it also has to move up I think
 
Why is the x coordinate of the turning point 1? What is the formula for the x coord of TP?
 
b^2 - 4ac? if it's smaller than zero, it's up in the air, is zero, it's on the x-axis, and if it's greater than zero, then it has 2 places where it hits the x-axis.

To get the turning point, i need to complete the square, but I need to know what k is.
 
The x-coord of the TP is b2a=32\displaystyle \frac{-b}{2a} = \frac {3}{2}.
 
Oh. I've seen (b/2)^2 for completing the square, but not that one I think.
 
So, now what is the y-coordinate of the TP? Hint: Put x=32\displaystyle x=\frac{3}{2} into the function and you will get an expression involving k.
 
Good. Now if the graph is shaped like a U and is totally above the x-axis (positive) then what do you know about the y-value of the TP?
 
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