What is the line tangent to the graph of #x=y^2+5# at the point (5,1)?

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Answer 1 · 2015-06-29T21:15:27

#y=1/2x-3/2#

#x=y^2+5#

#D(x)=D(y^2+5)#

#1=2y.y'+0#

#y'=(1)/(2y)#

Since #y=1# then:

#y'=(1)/(2xx1)=1/2#

The tangent is a straight line graph of the form:

#y=mx+c#

So at (5, 1):

#1=1/2x+c#

#1=5xx1/2+c#

#c=1-5/2=-3/2#

So the equation of the tangent #rArr#

#y=1/2x-3/2#