The parabola #f(x) = x^2 + 2(m+1)x + m + 3# is tangent to #x# axis in the negative side. What is the value of #m# ?

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Answer 1 · 2016-08-13T12:41:41

#m=1#

#f(x) = x^2+2(m+1)x+m+3#

#=(x+(m+1))^2-(m+1)^2+m+3#

#=(x+(m+1))^2-m^2-m+2#

This is in vertex form with intercept #-m^2-m+2#

In order for it to be tangential to the #x# axis the intercept must be #0#. That is:

#0 = -m^2-m+2 = (2+m)(1-m)#

So #m = -2# or #m = 1#.

If #m = -2# then the vertex is at #(1, 0)# on the positive part of the #x# axis.

If #m = 1# then the vertex is at #(-2, 0)# on the negative part of the #x# axis, as required.

#color(white)()#
Alternative method

#f(x) = x^2+2(m+1)x+m+3#

This is of the form #ax^2+bx+c# with #a=1#, #b=2(m+1)#, #c=(m+3)#

It has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (2(m+1))^2-4(1)(m+3)#

#=4(m^2+2m+1-m-3)#

#=4(m^2+m-2) = 4(m+2)(m-1)#

If the parabola is tangential to the #x# axis then it has a double root and we require #Delta = 0#.

Hence #m=-2# or #m=1#

If #m=-2# then #f(x)=x^2-2x+1 = (x-1)^2#, which is zero when #x=1#. This is on the positive part of the #x# axis.

If #m=1# then #f(x)=x^2+4x+4=(x+2)^2#, which is zero when #x=-2#. This is on the negative part of the #x# axis as required.