Quadratic equation help :)
Using complete sentences, explain how to find the average rate of change for f(x) from x = 4 to x = 7. The equation is y = 3(7-2)^2 + 98. I know that you start with f(4) = 3(4 - 2)^2 + 98 and f(7) = 3(7 - 2)^2 + 98, I'm just not sure where to go from here.
what is the function??
f(x) = 3(7-2)^2 + 98
that is not a function, that is just a number is there an \(x\) in it somewheres?
Sorry, I meant f(x) = 3(x-2)^2 + 98
ok what is \(f(7)\) ?
I'm not totally sure lol. Can we just start from scratch with f(x) = 3(x-2)^2 + 98?
sure
\[f(x)=3(x-2)^2+98\] and so \[f(7)=3(7-2)^2+98=3\times 5^2+98=3\times 25+98=75+98=176\]
in brief, \(f(7)=176\)
now you need \(f(4)\)
Okay, I'm still trying to figure out what you just did haha
where i saw an \(x\) i replaced it by a \(7\) and then computed the resulting number
I get replacing the x with 7, but I don't get = 3 × 52 + 98 = 3 × 25 + 98 = 75 + 98 = 176
should be \(3\times 5^2\) not \(3\times 52\)
\[3(7-2)^2+98\]What he did first was evaluate 7 - 2 = 5.\[3(5)^2+98\]Then he did 5^2=25.\[3(25)+98\]Then he did 3 * 25 = 75.\[75+98\]Which is 176.
what @bagajr said
Okay, I that makes sense. Only thing I don't get from that is the 5 (from 7 - 2 = 5)
what else could five minus two be?
I don't understand. Are you saying you don't understand why 7 minus 2 gives 5?
Wow I feel really stupid right now lol. I'm sorry, I wasn't paying attention at all. Never mind haha
That's alright, you must have just been overthinking things, lol
Yeah, we can go with that haha. Okay, so I understand all of that, so I'll try f(4) really quick.
\[4 - 2 = 2\] \[3(2)^2 + 98\] \[2^2 = 4\] \[3(4) + 98\] \[3 * 4 = 12\] \[12 + 98 = 110\] f(4) = 110? Is that right?
Yes
Alright cool! There's one more if you wouldn't mind :)
To find the rate of change between two points (in this case, x=4 and x=7), find the slope between them. Do you know how to do that?
*average rate of change
Oh yeah, forgot about that part lol. I think so, one sec.
I think it's 22?
Correct
Alright, thanks a lot! I might need help with the other, if I do, would you mind? :)
Okay, are you going to post it right here?
No, I'll ask another question then mention you, that way I can give you another medal lol
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