Following cordinate-points are given to define a function:
(˝,o), (-2,0) and (1,-3) with the parable roots being ˝ and -2. Since d>0, I used the formula: f(x)= a(x-r1)(x-r2), where r1= first root and r2= second root.
So this is what I end up with:
f(x) = a(x-˝)(x+2) <=> f(x)= a(x^2+ 2x- ˝x- 1) <=> f(x)= ax^2+ 1˝x-1
Then I isolate a, by inserting the x and y values of the third coordinate-point (1,-3) and I get:
-3= a(1-0.5)(1+2) <=> -3= a(1+2-0.5-1) <=> -3= a(1.5) <=> -3/1.5 = a <=> a= -2
So the final definition of the function is:
f(x) = -2ax^2 + 1˝x - 1
But in the answer list in my text-book, it says the definition should be:
f(x)= 2x^2 + 3x- 2
Can someone please explain to me where I went wrong?
Thank you in advance.
If a is the slope, then why does it appear thrice? The general formula for a quadratic is ax^2 + bx + c. But maybe it was just a typo? Or maybe there's something I'm just not seeing?
I'll be sure to rephrase 1˝x to 3/2x, but even so, it still doesn't solve my problem and i still end up with the same expression: f(x) = -2ax^2 + 3/2x - 1, which is somewhat of a far cry from what the answer in my textbook is, namely
f(x)= 2x^2 + 3x- 2
When you used the formula the a is out front so you need to expand it out. Indeed, it is not a slope, only straight lines have slope.
Do you understand that
Once simplified that is a factor of -1 out but I see no way to reconcile that (perhaps it's an even function?)
The value of f(x)= -2x^2 + 3/2x - 1 is not conserved if you rephrase the expression to f(x)= (-2*x^2) + (-2*1.5x) + (-2* -1), because the coefficient -2 is only to be multiplied with the quadratic term, as seen in the original expression I ended up with. Had the expression been f(x)= -2(x^2 +1.5x -1), then the above method would be right.
The only way I can see how you might have arrived at f(x)= -2(x^2 +1.5x -1) is if maybe you extracted it from the formula f(x)= a(x-r1)(x-r2). If so could you please show me how, step by step?
I apologise in advance for my limited math-abilities, I just want to understand, that's all.
Thanks for the contributions
Having to do it the other way around, i.e. breaking down a predefined functions into factors, made me understand what both of you were trying to say. Thanks for the help... And yeah I guess the problem is solved, well except the negative sign infront of a... Oh well maybe it's an error in the textbook.