# How do you evaluate the function f (x) = 3x^2 + 3x -2 for  f (2/a)?

Nov 20, 2015

By 'evaluate' I assume you mean find $a$ for $f \left(\frac{2}{a}\right) = 0$
$\textcolor{g r e e n}{a = - 1 \pm \sqrt{7}}$

#### Explanation:

Let $\textcolor{w h i t e}{\times} {y}_{1} = 3 {x}^{2} + 3 x - 2 \to f \left(x\right)$

Then ${y}_{1} = 3 {\left(\frac{2}{a}\right)}^{2} + 3 \left(\frac{2}{a}\right) - 2 \to f \left(\frac{2}{a}\right)$

${y}_{1} = 3 \left(\frac{4}{a} ^ 2\right) + 3 \left(\frac{2}{a}\right) - 2$

${y}_{1} = \frac{12}{a} ^ 2 + \frac{6}{a} - 2$

Let $\textcolor{w h i t e}{\times} {y}_{2} = \frac{1}{y} _ 1$ giving

${y}_{2} = \frac{1}{12} {a}^{2} + \frac{1}{6} a - \frac{1}{2} = 0$

Multiply throughout by 12 giving:

${y}_{2} = {a}^{2} + 2 a - 6 = 0$
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$\textcolor{g r e e n}{\text{Using std form} \textcolor{w h i t e}{\times x} a {x}^{2} + b x + c = 0}$

color(green)("Where" color(white)(xx)x = (-b+-sqrt(b^2-4ac))/(2a)

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In the context of the given question we have:

$a = \frac{- 2 \pm \sqrt{{\left(2\right)}^{2} - 4 \left(1\right) \left(- 6\right)}}{2 \left(1\right)}$

$a = \frac{- 2 \pm \sqrt{4 + 24}}{2}$

$a = - 1 \pm \sqrt{7}$

$a \approx - 1 \pm 2.65$ to 2 decimal places

$a \approx - 3.65 \textcolor{w h i t e}{x} : \textcolor{w h i t e}{x} a \approx 1.65$