Question

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

Answer 1

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`