At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Treat them as if you were completing the square
so\[x^2+6x-1=(x+3)^2-10\], the bracket sets them up looking like an a^2-10 (a=x+3), like any normal quadratic they start at a line of symmetry of x=0, so the translation of -10 means is irrelevant and the transformation +3 means the line of symmetry shifted to the left by 3 (i.e line of symmetry is x=-3)
Try it for the rest then you can compare their lines of symmetry
so for the 2nd one, take out the the negative. -x^2+2=-(x^2-2), line of symmetry is x=2
Yep! I understand now. Thank you so much!
Whoops, I've run into some trouble again. When comparing the line of symmetry for h(x), which I found to be positive 1, I realized it can't be the right answer. What am I doing wrong?
so factorising 2 out \[2x^2-4x+3=2(x^2-2x)+3=2(x-1)^2-1+3\] line of symmetry should be positive 1
oh after the 2 should still be 2(((x-1)^2)-1) but that still wouldnt make a difference to the line of symmetry
Exactly! However, according the practice module, that can't be right. Is there a possibility that f(x) is equivalent to positive 3 rather than negative as we had originally thought?
its still gonna be +3 in the brackets so thats a -3 translation, did you type up f(x) wrong?
Well, in the practice module, it gave these options for the ordering of the functions by line of symmetry (least to greatest):
A. f, g, h
B. h, g, f C. g, h, f D. h, f, g
so it should be in the order (small to greatest), f(x), h(x), g(x) :/
But our findings would suggest f, h, g, like you said, but it's not an option.
It must be an error in the module.
oh sorry i made a mistake with g
you dont need to bracket it because its -x^2, its an upside down quadratic so its symmetry is still 0 not 2
thus the order would be f, g, h?
Thank you for clearing that up! I was driving myself crazy. It's funny that I didn't notice the error in g(x) either.
it was my fault i thought -(x^2-2) was -(x-2)^2 XD