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1) what is the range of    y = 3 sqrt (x+6)

 

2) find the solution(s) to x+1 = sqrt(2x + 5) 

 

3) for the sequence 2/3 , 2/9 , 2/27 , 2/81 write the nth term formula     (Is it geometric or arithmetic? How do i know?)

 

4) find an equation for the inverse of the relation y = 9x+5

 

5) identify zeros and asymptotes of f(x) = (3x^2 + 2x - 8)/ (x^2 - 9)

 Jan 31, 2019
 #1
avatar+95866 
+2

1) what is the range of    y = 3 sqrt (x+6)

 

I believe this might be  y = ∛ (  x + 6)

 

As x assumes some large negative value........y tends toward - inf

As x assumes some large positive value......y tends toward inf

 

So   the range is (-inf, inf)

 

 

2) find the solution(s) to x+1 = sqrt(2x + 5) 

 

Square both sides

 

x^2 + 2x + 1 = 2x + 5       rearrange as

 

x^2 - 4 = 0         factor

 

(x + 2) (x - 2)  = 0

 

Setting each factor to 0 and solving for x we get that  x = -2    or  x = 2

 

The second solution is good

The first  is extraneous.....it makes the left side of the original equation negative....but we cannot get a negative from a positive radical

 

 

cool cool cool

 Jan 31, 2019
 #2
avatar+95866 
+2

3) for the sequence 2/3 , 2/9 , 2/27 , 2/81 write the nth term formula     (Is it geometric or arithmetic? How do i know?)

 

Geometric....the common ratio is  1/3

 

Arithmetic sequences hve a common difference between terms, not a common ratio between terms

 

The nth term formula is     (2/3)(1/3)^(n - 1)

 

 

4.   y = 9x + 5       subtract 5 from both sides

 

y - 5 =  9x          divide both sides by  9

 

[ y - 5  ] / 9 = x            "swap" x and y

 

[ x - 5 ] / 9 = y          the inverse   is in red

 

 

 

 

cool cool cool

 Jan 31, 2019
 #3
avatar+95866 
+2

5) identify zeros and asymptotes of f(x) = (3x^2 + 2x - 8)/ (x^2 - 9)

 

The numerator determines the zeroes

 

3x^2 + 2x - 8 = 0

 

(3x - 4) (x + 2) = 0

Setting each factor to 0 and solving for x gives the zeroes of x = 4/3  and x = -2

 

We have same power polynomial / same power polynomial    situation

The horizontal asymptote comes from the ratio of the coefficients on the x^2 terms

This is

y = 3/1   =  3

 

 

The vertical asymptotes are the x values that make the denoninator = 0

 

Note that these are  x = 3      and x = -3

 

 

cool cool cool

 Jan 31, 2019

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