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# Given that the graphs of \$y=h(x)\$ and \$y=j(x)\$ intersect at \$(2,2),\$ \$(4,6),\$ \$(6,12),\$ and \$(8,12),\$ there is one point where the graphs of

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Given that the graphs of \$y=h(x)\$ and \$y=j(x)\$ intersect at \$(2,2),\$ \$(4,6),\$ \$(6,12),\$ and \$(8,12),\$ there is one point where the graphs of \$y=h(2x)\$ and \$y=2j(x)\$ must intersect. What is the sum of the coordinates of that point?

Aug 28, 2017

#1
+8194
+5

The given intersection points lie on  y = h(x)  and  y = j(x)  .

For example...since  (8, 12)  is an intersection point....   12 = h(8)   and   12 = j(8)

When  x = 4  ,     h(2x)    =    h(2(4))    =    h(8)    =    12    =    y

When  x = 4  ,     2j(x)     =    2j(4)       =    2(6)    =    12    =    y

So... (4, 12)  is a point on  y = h(2x)  and on  y = 2j(x)  .

4 + 12  =  16

Sorry that this isn't a very good explanation.....If it doesn't make sense, it might help to try different numbers for  x  , such as 1, 2, or 6...to see why they don't work.

Aug 29, 2017

#1
+8194
+5

The given intersection points lie on  y = h(x)  and  y = j(x)  .

For example...since  (8, 12)  is an intersection point....   12 = h(8)   and   12 = j(8)

When  x = 4  ,     h(2x)    =    h(2(4))    =    h(8)    =    12    =    y

When  x = 4  ,     2j(x)     =    2j(4)       =    2(6)    =    12    =    y

So... (4, 12)  is a point on  y = h(2x)  and on  y = 2j(x)  .

4 + 12  =  16

Sorry that this isn't a very good explanation.....If it doesn't make sense, it might help to try different numbers for  x  , such as 1, 2, or 6...to see why they don't work.

hectictar Aug 29, 2017
#2
+101360
+1

Thanks, hectictar......these often throw me.....!!!!

Aug 29, 2017