Find the constant a so that the graph of y=1/(ax+2) has tangent
y=4y+3x-2=0 at the point (0,1/2)
Please show work.
Thanks in advance!
y=4y+3x-2=0 at the point (0,1/2)
Please show work.
Thanks in advance!
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y = 4y + 3x - 2 = 0
y = (2 - 3x)/4 -->m = -3/4
y' = -a/(ax + 2)²
-a/4 = -3/4
a = 3
y = (2 - 3x)/4 -->m = -3/4
y' = -a/(ax + 2)²
-a/4 = -3/4
a = 3
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@chris: it is if you get the y's to one side then divide by 3
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"y=4y+3x-2=0" is not a tangent line.