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

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

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Answer 1 · 2016-07-30T18:39:09

I assume that $(f o g) (x)$ $(fog)(x)$ means $f (g (x))$ $f(g(x))$

Explanation:

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

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

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

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

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

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

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

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If f(x)=2x2+5f(x) = 2x^2 + 5 and g(x)=3x+ag(x) = 3x + a, how do you find a so that the graph of (f o g)(x) crosses the y-axis at 23?

1 Answer

Explanation:

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