How do you find the x intercepts for #y=16+5x+(3/x)#?

1 Answer
David G.
Jun 15, 2016

The x-intercepts are the points on the curve where #y=0#. Solving the equation #3/x+5x+16=0# yields the x values #-3# and #-1/5#, so the coordinates of the points are #(-3,0) # and #(-1/5,0) #.

Explanation:

#3/x+5x+16=0#

Multiply through by #x#:

#3+5x^2+16x=0#

And now we have a quadratic equation, and we know how to solve those. There are several methods, including factorisation, but my favourite is the quadratic formula.

Rearranging:

#5x^2+16x+3=0#

The quadratic formula applies to a quadratic equation in the form #ax^2+bx+c=0#, and goes:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

In this case,

#x=(-16+-sqrt(16^2-4*5*3))/(2*5)=(-16+-sqrt(256-60))/(10)#

#=(-16+-14)/(10) =-3 or -1/5#

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