CPhill

Let's see if this helps......."b" shows how far the basic graph of y = lxl is shifted either left or right. Since the "vertex" of the basic graph is at (0, 0), the "vertex" of this graph lies at (6, -1). Thus, the graph is shifted to the right by 6 units, so b = 6. "a" actually isn't the slope, it serves to either widen or narrow the basic graph (this one is narrower)....well anyway, we can find "a" thusly:

We have the point on the graph (0, 11)....so

y = alx - 6l - 1     and filling in for x and y, we have

11 = al0 - 6l - 1   add 1 to both sides

12 = al-6l     

12 = a*6       divide by 6 on both sides

2 = a

Does that help??

So I guess you want to graph y = l 7 - x l ???

"y"  will never be negative in this graph.

You have the first part correct.  When x = 0, y will = l 7 l  = 7

So we have the points   (7, 0) and (0, 7)

Connect these two points. Now, pick a value for x > 7, say 9. Then l 9 - 7 l = 2. So the point (9, 2) will be on the graph. Draw a line connecting this point and (7, 0). There's your graph.

The easiest thing to see is that this is the graph of y = lxl shifted to the "right" by 7 units ....!!!!     y = lx + 7l would shift the basic graph to the "left" by 7 units ....

Does that help??

Well...if I don't know much about them, someone on here probably does.....post your questions......

LOL!!!...don't worry about it......remember, you're "Anonymous"......

We don't need much familiarity with logs or exponentials to do this one. We're just asking, "What exponent do I put on 2 to equal 2??"  The answer is 1. This is because 2^1 = 2 ....Thus, x = 1......

Here's my approach to (c)

lim x → ∞  [sinx^2 + 10] /[x^ + 10]

Splitting this up, we have

lim x → ∞  [sinx^2 ] / [x^ + 10]  + lim x → ∞  [ 10] / [x^ + 10]

Notice that the second thing just evaluates to 0

For the first...we can do the Squeeze Theorem "thing" again

-1 ≤ sin x^2 ≤ 1

And dividing each thing by x^2 + 10, we have

-1 / x^2 + 10 ≤ sin x^2 / x^2 + 10  ≤ 1 / X^2 + 10

And because the two "outside" functions approach 0 as x approaches infinity, then the middle one is "squeezed" to this limit, too.

Thus

lim x → ∞  [sinx^2 ] / [x^ + 10]  + lim x → ∞  [ 10] / [x^ + 10]  =

                  0                              +               0   =                                0         !!!!

Thanks for that excellent answer, Melody.........I hope "Anonymous" understood it better after that...

Here's why 0^0 is undefined.....

Normally  a^n /a^n = a^(n-n) = a^0 =1

So....   0^a/0^a  should be  0^(a - a) = 0^0 =1....however, the problem is that we divided by 0 originally (0^a)....and that's undefined

I assume we have

log₁₆ 19     using the "change of base" rule we have

log19/log16 = about 1.062

Very nice !!!!....you're not as "limited" as you think !!!!