HELPPPPPPPPP!!!!!!!!!!!!!! For the graphed function f(x) = (2)x + 2 + 1, calculate the average rate of change from x = -2 to x = 0. graph of f of x equals 2 to the x plus 2 power, plus 1. −3 over 2 −2 3 over 2 4
do you understand the idea of average rate of change?
no
average rate of change = f(2) - f(1) / b - a
how would i work that out with the graph?
sorry... f(b) - f(a) / b - a
I'm still lost
think of the average rate of change as slope
Still lost :(
Thats the graph
There's something wrong, when you plug in 0 for x into 2x + 2 + 1 it equals 0 + 2 + 1 = 3 (0,3), your graph says (0,5). You said, "2 to the x" are you implying 2^x? Because your graph looks to be exponential. In my previous attempt I mistook it for a quadratic graph.
I'm not sure,thats the way I was given
IF it was exponential it would actually give you (0,4) because 2^0 = 1, 1 + 3 = 4
thank youuuu, I kinda of get it now...,can you help me on a few other graph questions as well?
"Graph of f of x equals \(\ \sf 2 ~ to~ the ~x \) plus \(\ \sf 2 ~ POWER\), plus 1. What do you mean "2 power"?
Is it like an exponent?
I've never seen a problem say "2 power". Can you screen shot the entire problem along with the graph?
okay
Exactly ^_^ There's a big difference between 2x + 2 + 1 and \(\ \sf 2^{x + 2} + 1 \) Can you plug in -2 into f(x) = \(\ \sf 2^{x + 2} + 1 \) for me?
Just replace "X" with -2 everywhere there is an "x"
one sec
k
is it -2?
\(\ \sf 2^{-2 + 2} + 1 = 2^0 + 1 = 1 + 1 = 2 \) So yes, (-2,2) is correct. Now do 0?
4?
Close, \(\ \sf 2{0 + 2} + 1 = 2^2 + 1 = 4 + 1 = 5 \) So (0,5)
what would I have to do next?
Find the average of change between x = -2 to x = 0, The formula I'm familiar with is \(\ \sf \dfrac{f(b) - f(a)}{b - a} \) \(\ \sf \dfrac{5 - 2}{0 - (-2)} = 3/2 \) I also did this in the previous problem and got 3/1, was it wrong?
3/2*
It was right,thank you!!I have 4 more graph questions I need some help on please
Brb!
graph begins in the second quadrant near the line y equals 3 and increases slowly while crossing the ordered pair 0, 4. When the graph enters the first quadrant, it begins to increase quickly throughout the graph. f(x) = 4x f(x) = 4x − 3 f(x) = 4x + 3 f(x) = 4(x + 3)
ok
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