f(x) = 4- sqrt (16 + 6x - x^2)

Nice!

Now I get it!

Thanks a lot!

You have any idea how one would find Ima f(x) ?

Original answer by geno3141:

f(x)  =  4 - sqrt(16 + 6x - x2)

For ease of calculation, I am replacing  f(x)  with  y:

                                                                  y  =  4 - sqrt(16 + 6x - x2)

Subtract 4 from both sides:                  y - 4  =  - sqrt(16 + 6x - x2)

Multiply both sides by -1:                    -y + 4  =  sqrt(16 + 6x - x2)

Simplify left side:                                   4 - y  =  sqrt(16 + 6x - x2)

Square both sides:                            (4 - y)2  =  16 + 6x - x2

Multiply out:                                16 - 8y + y2  =  16 + 6x - x2

Subtract 16 from both sides:           - 8y + y2  =  6x - x2

Rewrite:                                              y2 - 8y  =  -x2 + 6x

Factor:                                                y2 - 8y  =  -(x2 - 6x)

Complete the squares:    [ y2 - 8y + 16 ] - 16  =  [ -(x2 - 6x + 9) ] + 9

Simplify:                                   [ y2 - 8y + 16 ] =  [ -(x2 - 6x + 9) ] + 25

Factor:                                               (y + 4)2  =  -(x - 3)2 + 25

Rewrite:                              (x - 3)2 + (y + 4)2  =  25  

Analysis:  this is the equation of a circle with center  (3, -4)  and  radius = 5.

But:  the initial equation is only the lower half of this circle; thus, an arc from  (-2,4)  through  (3,-1)  to  8,4).

You can use Desmos and copy the hyperlink to your  post...... [the website calculator is "clunky," IMHO !!!!  ]

Here's the graph of this function : https://www.desmos.com/calculator/s3u8kl73w3

f(x) =   4- sqrt (16 + 6x - x^2)

Notice that the term inside the square root is only valid when its ≥  0

So......solving this we have

16 + 6x - x^2 ≥  0    factor

(-x + 8) (x + 2)  ≥ 0

Note that this is only true  wnen  x ≥ -2  and x ≤ 8

So....the domain of the function is   [-2, 8 ]

Note to find the intersection with the y axis,, set x = 0, the term inside the square root will equal 16 ...   so  y =    4 - √16 =   4 - 4  = 0 ........so when x = 0, then y = 0 

This is confirmed by the graph.....it contains the point (0, 0)......so it intersects the origin

And the other intersection point with the x axis  is just determined by setting hte function to 0 ....so we have

0  = 4 -  sqrt (16 + 6x - x^2)   rearrange

sqrt (16 + 6x - x^2)  = 4     square both sides

16 + 6x - x^2  = 16       subtract 16 from both sides

-x^2 + 6x  = 0       factor

x ( -x + 6)  = 0     set both factors to 0   and we have that the intersection points with the x axis occur at x = 0  [ which we already know ]   and  x = 6

Is that all you neede to know???

x(-x+6)=0

Setting both factors  to 0 , we have

x = 0         and      -x + 6 = 0

The first is obvious

For the second, we have

-x + 6 = 0     add x to both sides

6 = x

These are the two x intercepts  at x = 0 and x = 6

Remember that, at an x intercept, y = 0

So......if we plug 0 into the function, we get

y = 4 - sqrt [16  + 6(0)  - 0^2 ]  =

4 - sqrt (16)  =

4 - 4  = 0

This shows that x = 0 must be a root  [ i.e., an x intercept ] because when we plug this value into the function we get y = 0 back

Likewise....putting  x = 6 into the function we have

y = 4  - sqrt [ 16 + 6(6)  - 6^2 ]  =

4  - sqrt [ 16 + 36 - 36 ] =

4 - sqrt [ 16]  =    

4 - 4 = 0     so    x = 6 must also be a root [ i.e., an x intercept ]

Remember something else......if  a * b   = 0    then either a = 0  or b = 0   or they both equal 0

So....if we have   x ( -x + 6)  = 0 , then either  x = 0  or  -x + 6 =0 

And we needed to solve both to find out what makes each factor equal to 0

Sorry, Tony.....I'm  not familiar with images of a function....

Here's a resource that might help you.....

There does not seem to be any link there haha

Oops...sorry......LOL!!!

Here it is....http://pennance.us/home/downloads/3018/image.pdf

If I can figure this out.....I'll see if I can help you....