Question

Algebra II help, please?

Answer 1

Two athletes, Jack and Jill, invite you to participate in a 5K run with several hills. One hill can be represented by the function f(x) = 2x2 + 5. Another hill can be represented by the function g(x) = 2x2 - 3x + 4. Describe to Jack and Jill, using complete sentences, which of the operations-addition, subtraction, multiplication, and division-will result in the largest degree function and which operation will result in the smallest degree function.

Answer 2

@johnweldon1993

Answer 3

So...when you multiply sayyy

\[\large x^2 \times x^2\]

what do you get?

Answer 4

x^4

Answer 5

Right...notice how the degree (exponent) went up?

So if you add or subtract sayyy

\[\large x^2 + 3x^2\]
\[\large x^2 - 3x^2\]

you would get something with the same degree of 2 in both casescorrect?

Answer 6

I don't think so..!

Answer 7

Wait. Yes?

Answer 8

Yes indeed...when adding or subtracting something with the same variable and exponent...you leave the variable and exponent alone and just add or subtract the numbers.

So finally...if you divide

\[\large \frac{4x^2}{2x^2} = ?\]

Answer 9

2x^2?

Answer 10

Not quite...the x^2's now cancel

\[\large \frac{4\cancel{x^2}}{2\cancel{x^2}} = 2\]

if you have the x^2 on both top and bottom...they cancel! make sense?

Answer 11

Ohhh right!

Answer 12

Alright so, out of those 4 operations

which gave the highest degree?

Answer 13

Multiplication?

Answer 14

Correct.....when you multiply variables with exponents...the exponents add....so they almost always make a higher degree

And what about the lowest?

Answer 15

Division!

Answer 16

Correct :)

Answer 17

Great! But how would I prove this with the given problem?

Answer 18

So we just take the 2 equations

\[\large (2x^2 + 5) \times (2x^2 - 3x + 4)\]
What would be the result from that?

Answer 19

4x^4-6x^3+18x^2-15x+20

Answer 20

Perfect

And the division part

\[\large \frac{2x^2 + 5}{2x^2 - 3x + 4} = ?\]

Answer 21

\[\frac{ (2x^2+5) }{ (2x^2-3x+4) }\]

Answer 22

@johnweldon1993 ?

Answer 23

Actually I think I can go from here. Thank you so very much!