If y varies inversely as the square of x and y = 8 when x = 2, then what is the value of y when x = -1?

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Answer 1 · 2016-04-15T10:55:20

The answer is $32$ $32$ , based on an inverse proportionality model.

According to the statement,

$y \propto \frac{1}{x^{2}}$ $y prop 1/x^2$

This expression takes us to the exact formula:

$y = \frac{α}{x^{2}}$ $y = alpha / x^2$

where $α$ $alpha$ is a constant of proportionality. To find it, we need a numerical value: $y (x = 2) = 8$ $y(x=2) = 8$ . By substituting:

$8 = \frac{α}{2^{2}} \to α = 32$ $8 = alpha / 2^2 rightarrow alpha = 32$

So our formula is:

$y = \frac{32}{x^{2}}$ $y = 32 / x^2$

Now, let us calculate $y (x = - 1)$ $y(x=-1)$ :

$y = \frac{32}{{(- 1)}^{2}} = 32$ $y = 32 / (-1)^2 = 32$