# If f(x) = 2x^2 + 5 and g(x) = 3x + a, how do you find a so that the graph of (f o g)(x) crosses the y-axis at 23?

Jul 30, 2016

I assume that $\left(f o g\right) \left(x\right)$ means $f \left(g \left(x\right)\right)$

#### Explanation:

First let's find $f \left(g \left(x\right)\right) = 2 {\left(3 x + a\right)}^{2} + 5$, replacing $\left(3 x + a\right)$ into $f \left(\right)$. We then have $f \left(g \left(x\right)\right) = 2 \left(9 {x}^{2} + 6 a + {a}^{2}\right) + 5 = \left(18 {x}^{2} + 12 a + 2 {a}^{2}\right) + 5$.

Now the graph crosses the y-axis when $x = 0$, so we must have:

$12 a + 2 {a}^{2} + 5 = 23$, that is $12 a + 2 {a}^{2} - 18 = 2 {a}^{2} + 12 a - 18 = 0$. We must now solve the equation for $a$.

We may first divide the equation by $2$ (just to make the calculation slightly easier), so we get ${a}^{2} + 6 a - 9 = 0$, then the roots of the quadratic equation are:

$\frac{- 6 \pm \sqrt{36 + 4 \cdot 9}}{2} = \frac{- 6 \pm \sqrt{72}}{2} = \frac{- 6 \pm 6 \sqrt{2}}{2}$.

Then the two solutions are $a = \left(- 3 + 3 \sqrt{2}\right)$ and $a = \left(- 3 - 3 \sqrt{2}\right)$