Functions

Sophie02

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How would you go about doing part c? As you can see from the working out, my range for a is defiantly wrong and I don’t know how else to go about it?
 

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Jomo

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(2x+3)/x is NOT 3.
2x/x is NOT 0, and even if it was 0, you still need to divide the 3 by x!

(2x+3)/x = (2x/x) + (3/x) = 2 + 3/x.

Think about what is required for the line y=ax (a line that crosses the origin) to cross the function in exactly one place. Do not use algebra, but rather think how this can be done. Draw lines and figure out the requirement for the slope.
 
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Sophie02

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(2x+3)/x is NOT 3. 2x/x is NOT 0, and even if it was 0, you still need to divide the 3 by x!

(2x+3)/x = (2x/x) + (3/x) = 2 + 3/x.

Think about what is required for the line y=ax (a line that crosses the origin) to cross the function in exactly one place. Do not use algebra, but rather think how this can be done. Draw lines and figure out the requirement for the slope.
Is the answer a>2 and a<-2?
 

Jomo

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Sophie02

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Maybe? Can you explain why? Did you miss one other a value?
I used desmos to see where the lines would cross. But it is to do with how steep the gradients are. But I don’t think there is another one because it seems to cross both lines otherwise
 

HallsofIvy

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You say your answer to (a) is "defiantly wrong" but you don't want help with that?

You have the graph of y= 2|x+ 4|- 5. The best way to handle an absolute problem is to look at "cases".

1) if x> -4, x+ 4 is positive so |x+ 4|= x+ 4. y= 2|x+ 4|- 5= 2(x+ 4)- 5= 2x+ 8- 5= 2x+ 3.

2) If x< -4, x+ 4 is negative so |x+ 4|= -(x+ 4)= -x- 4. y= 2|x+ 4|- 5= 2(-x- 4)- 5= -2x- 8- 5= -2x- 13.

(a) P is where the two lines intersect, at x=-4. y= 2(-4)+ 3= -5 and y= -2(-4)- 13= 8- 13= -5.
P= (-4, -5).

(b) says "solve the equation". The only equation here is y= 2|x+ 4|- 5 so I would conclude that this asking us to solve for x in terms of y. Again,write in two cases:
y= 2x+ 3 so y- 3= 2x and x= (y- 3)/2. If x> -4 then y> -5.
y=-2x- 13 so y+ 13= -2x and x= -(y+ 13)/2. If x< -4 then y< -5

so x= (y- 3)/2 for y>-5 and x= -(y+ 13)/2 for y< -5.

For (c) we have a line y= ax. The right side of the absolute value function has slope 2. y= ax has slope a and will be parallel to that line for a= 2. Any a< 2 will intersect at a point above the x-axis. The slope of the line from (-4, 5) to (0, 0) is -5/4 so any a> -5/4 will intersect below the x-axis. a must be between -2 and -5/4.
 

Jomo

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Is the answer a>2 and a<-2?
No, a>2 and a<2 is not part of the answer. Find the slope of each line, and then see which lines through the origin will cross just one of the two lines. Then there is one more line.
 
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