By mcgowan04 - 5/27/2005 6:58:24 PM
How are you approaching Exercise 1.2? I was thinking NPV rule or rate of return Rule, but I think that's too much. For part A are we just supposed to pick 1 project for each firm? Wouldn't that just be the project with the highest yield? For part B I dont even know where to begin.
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By sheilaaellis44 - 3/27/2012 12:39:16 AM
I will try my best to help you.
y=a(x+2.4)(x-0.8) <-good job!
Now you know the parabola passes through a point(her coordinate is (1.2,–2.592) as you know).This is means even if we haven't known the letter "a" equals what,but we can conclude that where "x=1.2",it must exist"y=-2.592".So we can let x=1.2,y=-2.592 in equation to figure out the value of "a".
y=a(x+2.4)(x-0.8)
–2.592=a(1.2+2.4)(1.2-0.8)
So a=-1.8
then y=-1.8(x+2.4)(x-0.8)=3.456-2.88x-1.8x^2(… the factored form)
Please check the answer by yourself,it was calculated and expanded by mathematical software named"Mathematica".I am not responsible for its accuracy.
If you know a quadratic function just like this form:y=ax^2+bx+c (while the damned "a"doesn't equals zero;a,b or c are constants),you can know that:the x-coordinate of the vertex is -b/(2a),and the y-coordinate is (4ac-b^2)/4a.The conclusion can be inference by completing square.Every student have to recite it.
In this example“y=-1.8x^2-2.88x+3.456”,you can get it that"a=-1.8,b=-2.88,c=3.456".Then the x-coordinate of the vertex is -b/(2a)=-(-2.88)/(2×(-1.8))=-0.8,the y-coordinate of the vertax is (4ac-b^2)/4a=4.608(leave its process out for convinient).
According to "the point on the x-axis halfway between the two x-intercepts",you can also do "(-2.4+0.8)/2=-0.8" to calculated x-coordinate of the vertex.
vertex form:y=a(x+b/(2a))^2+(4ac-b^2)/4a
in your example is y=-1.8(x+0.8)^2+4.608
There is some calculated mistake in my software(maybe),please check the answers by yourself.
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