Question

how to find the range of f(x)=2+sqrt(8x-2x^2)

Answer 1

the square root of something is never negative, so the lower bound is \(y=2\)

Answer 2

I want to see how to do it. My Teacher showed us different way of doing it

Answer 3

to find the upper bound, find the maximum value of \(8x-x^2\)

Answer 4

you know how to find that? it is the same as finding the second coordinate of the vertex

Answer 5

can you show me?I want to make sure I am on the right track. please

Answer 6

I got 2< or= f(x)<2+(sqrt.(8))

Answer 7

first coordinate of the vertex is always \(-\frac{b}{2a}\) which in your case is \[-\frac{8}{2\times -2}=2\]

Answer 8

second coordinate is what you get when you replace \(x\) by \(2\) namely 8
so your answer
\[2\leq y\leq 2+\sqrt{8}\] is correct

Answer 9

can you show me how will you get this one...because you may have an easier solution and easier to understand solution

Answer 10

what is it?

Answer 11

your solution of how you will get this

Answer 12

probably the same way you did it, since we got the same answer
i wrote my method above
find the second coordinate of the vertex of \(8x-2x^2\)

Answer 13

this is how my professor did it:
\[f(x)\ge2\]
\[y=8x-2x ^{2}\]
then he made a table of values...
Is there an algebraic way of doing it?

Answer 14

yes
read the answer i wrote above

Answer 15

can I see the solution please...I kinda get it

Answer 16

i can write it again if you like, but i wrote it above
the first coordinate of the vertex is \(8x-2x^2\) is \(-\frac{b}{2a}=-\frac{8}{2\times (-2)}=2\)

Answer 17

the second coordinate of the vertex is \(8\times 2-2\times 2^2=8\)

Answer 18

so the max of your function is \(2+\sqrt{8}\)

Answer 19

Oh I get it...Thanks

Answer 20

yw

Answer 21

do I always have to use the equation under the root to find the vertex?

Answer 22

depends on the function

Answer 23

where did you get -b/2a?