Question #e07a4

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Question #e07a4

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Answer 1 · 2016-08-31T17:24:17

$(f + g) (- 2) = 24$ $(f+g)(-2)=24$

Explanation:

Using the notation that for given functions $f$ $f$ and $g$ $g$ that $(f + g) (x) = f (x) + g (x)$ $(f+g)(x) = f(x) + g(x)$ , we have

$(f + g) (- 2) = f (- 2) + g (- 2)$ $(f+g)(-2) = f(-2)+g(-2)$

$= (2 {(- 2)}^{2} - 7 (- 2) + 6) + ((- 2) - 2)$ $=(2(-2)^2-7(-2)+6) + ((-2)-2)$

$= (8 + 14 + 6) + (- 4)$ $=(8+14+6)+(-4)$

$= 30 - 4$ $=30-4$

$= 24$ $=24$

Note that we can also solve generally for $(f + g) (x)$ $(f+g)(x)$ and then plug $- 2$ $-2$ in:

$(f + g) (x) = f (x) + g (x)$ $(f+g)(x) = f(x) + g(x)$

$= (2 x^{2} - 7 x + 6) + (x - 2)$ $= (2x^2-7x+6) + (x-2)$

$= 2 x^{2} - 6 x + 4$ $=2x^2-6x+4$

Plugging in $- 2$ $-2$ leads to the same result:

$2 {(- 2)}^{2} - 6 (- 2) + 4 = 8 + 12 + 4 = 24$ $2(-2)^2-6(-2)+4 = 8 + 12 + 4 = 24$

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Question #e07a4

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