How do you find the x intercepts for $y = 16 + 5 x + (\frac{3}{x})$ $y=16+5x+(3/x)$ ?

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How do you find the x intercepts for $y = 16 + 5 x + (\frac{3}{x})$ $y=16+5x+(3/x)$ ?

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Answer 1 · 2016-06-15T21:26:16

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

Explanation:

$\frac{3}{x} + 5 x + 16 = 0$ $3/x+5x+16=0$

Multiply through by $x$ $x$ :

$3 + 5 x^{2} + 16 x = 0$ $3+5x^2+16x=0$

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:

$5 x^{2} + 16 x + 3 = 0$ $5x^2+16x+3=0$

The quadratic formula applies to a quadratic equation in the form $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , and goes:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case,

$x = \frac{- 16 \pm \sqrt{16^{2} - 4 \cdot 5 \cdot 3}}{2 \cdot 5} = \frac{- 16 \pm \sqrt{256 - 60}}{10}$ $x=(-16+-sqrt(16^2-4*5*3))/(2*5)=(-16+-sqrt(256-60))/(10)$

$= \frac{- 16 \pm 14}{10} = - 3 or - \frac{1}{5}$ $=(-16+-14)/(10) =-3 or -1/5$

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How do you find the x intercepts for $y = 16 + 5 x + (\frac{3}{x})$ $y=16+5x+(3/x)$ ?

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