Let t(x)=-2x^3+5x-7. Find the new function for t(x) under the following transformations. Express t(x) as a cubic polynomial in the form ax^3+bx^2+cx+d. A. Shift t(x) up 7 units and left 5 units B.Reflect t(x) across the y-axis c. Stretch t(x) horizontally so that it is twice as wide as the original function.
please help lol
to shift the function up 7 units add 7 to t(x) the rule to shift a function a units to the left is f(x) ----> f(x+a) so for t(x) you replace each each x by x + 5 and simplify
So i'll start you off with A:- adding 7 we get -2x^3 + 5x - 7 + 7 = -2x^3 + 5x Now replace the x by x + 5 and express it in the form required in the question.
2(x+5)^3+5(x+5). =2x^3+30x^2+155x+275
I just simplified it
looks good
is that the answer?
thats the answer to A
B when x is reflected acros the y axis the y values stay the same but the x values change to -x. So if you replace each x by -x in -2x^3 + 5x - 7 you have the answer to B
* when f(x) is reflected
- I should have said t(x)!
so its just 2x^3-5x-7?
yeah
C:- for a horizontal stretch of twice the width you multiply the whole function by 1/2
the original function correct?
OH Hold on I'm not sure about that I'll check that out.
oh ok lol
I was wrong you need to replace the x by 0.5x then simplify
so we have -2(0.5x)^3 + 5(0.5x) - 7
lol my brains dead. been up since 11 at night. ok so -.25^3+2.5x-7?
lol sorry You are correct
oh lol so is that it?
Yes I have assumed that the 3 transformations are done on the original t(x) . Not consecutive. The question does not make that clear.
oh ok thank you so much. you mind helping me out on organizing the quadratic formula?
ok
12x+13-5x^2 I have it as -12+-sqrt12^2-4(-5)13 all over 2(-5)
that's correct
yes however I cannot seem to get the answer.
I just want to simplify it
work out the stuff thats in the brackets first 12^2 - 4*-5*13 = 144 + 260 = 404
so we have (-12 +/- sqrt 404 ) / -10
perhaps you may assist someone else right now. I have to go to the restroom for an emergency lol ill be right back
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