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f(x)=-20x²+14x+12 and g(x)=5x-6

 May 6, 2014
 #1
avatar+118667 
0

Hi All SmilesAlways,

Since you are dividing by 5x-6,

 $$5x-5\ne 0 \rightarrow x\ne 1$$

 

$$f/g=\frac{-20x^2+14x+12}{5x-6}\\\\
f/g=\frac{-2(10x^2+7x+6)}{5x-6}\\\\
f/g=\frac{-2(2x+1)(5x-6)}{5x-6}\\\\
f/g=-2(2x+1)$$

This is a line with a hole at (1,-6)

Domain All real x where x is not equal to 1

Range: All real y where y is not equal to -6

I am very tired so you need to check my working. 

 May 6, 2014
 #2
avatar+2353 
0

Sorry Melody, you may want to adjust some things;

 

$$5x-6 \neq 0$$

$$x \neq \frac{6}{5}$$

 

and

$$(2x+1)(5x-6) = 10x^2+5x-12x-6$$

which is

$$10x^2-7x-6$$

but since $$-20x^2+14x+6 = -2(10x^2-7x-6)$$

The rest does work.

Nevertheless the hole is at x= 6/5

 May 6, 2014
 #3
avatar+118667 
0

Thanks reinout,

Reinout is right.  I think mine is ok except the hole is in the wrong place.

therefore the range and domain are not correct.

Reinout could you please give AllSmilesAlways the correct answer.

I have had enough for the night.  It is 2am.

See you much later in the day. 

 May 6, 2014

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