SOLUTION: Let f(x) = 2x^2 - 3x +4 and g(x) = x + 3. Find f(x) + 3g(x) A. 2x^2 - 2x + 7 B. 2x^2 + 6x + 13 C. 2x^2 + 13 D. 2x^2

SOLUTION: Let f(x) = 2x^2 - 3x +4 and g(x) = x + 3. Find f(x) + 3g(x) A. 2x^2 - 2x + 7 B. 2x^2 + 6x + 13 C. 2x^2 + 13 D. 2x^2 - 6x + 13 2. State the range

Question 762286: Let f(x) = 2x^2 - 3x +4 and g(x) = x + 3.
Find f(x) + 3g(x)

A. 2x^2 - 2x + 7

B. 2x^2 + 6x + 13

C. 2x^2 + 13

D. 2x^2 - 6x + 13

2.
State the range of the following relation:
{(-1, -3), (9, -1), (-6, -9), (0, -4), (2, -1), (-2, -3)}

A. {All real numbers}

B. {-9, -4, -3, -1}

C. {-1, -3, 9, -6, 0}

D. {-6, -2, -1, 0, 2, 9}
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
# 1

f(x) + 3*g(x) = 2x^2 - 3x +4 + 3*(x+3)

f(x) + 3*g(x) = 2x^2 - 3x +4 + 3x + 9

f(x) + 3*g(x) = 2x^2 + 13

So it's choice C

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# 2

The range is the set of all possible outputs of a function. Basically it's the set of y values of each point.

The y values of (-1, -3), (9, -1), (-6, -9), (0, -4), (2, -1), (-2, -3) are: -3, -1, -9, -4, -1, -3

Toss out the duplicates to get: -3, -1, -9, -4

and this is choice B (just in a different order, but order doesn't matter in a set)
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