Question

question: 13
which of the following describes the interval of increase for the graphed function (green function)?
a. $(-\infty, \infty)$
b. $(-\infty, 3)$
c. $(3, \infty)$
d. $(1, \infty)$

question: 14
what are the zeros of the function $g(x) = x^2 + 4x - 12$
a. $g(0) = 2$ and $g(0) = -6$
b. $g(0) = -2$ and $g(0) = 6$
c. $g(-2) = 0$ and $g(6) = 0$
d. $g(2) = 0$ and $g(-6) = 0$

Explanation:

Question 13 (Assuming the green function is a linear function, as option A is $(-\infty, \infty)$ which is the interval of increase for a linear function with positive slope, or a function that's always increasing. But since the options include $(-\infty, \infty)$, we'll solve based on that logic.)

Step1: Recall interval of increase

A function is increasing on an interval if, as \( x \) increases, \( f(x) \) also increases. For a linear function (like \( y = mx + b \) with \( m>0 \)) or a function that's always increasing (e.g., exponential with positive base), the interval of increase is \( (-\infty, \infty) \).

Step2: Analyze options

• Option A: \( (-\infty, \infty) \) suggests the function is always increasing.
• Option B: \( (-\infty, 3) \) suggests increase only left of \( x=3 \), then maybe decrease.
• Option C: \( (3, \infty) \) suggests increase only right of \( x=3 \).
• Option D: \( (1, \infty) \) suggests increase only right of \( x=1 \).

If the green function is linear (e.g., a straight line with positive slope) or a function that never decreases, the interval of increase is \( (-\infty, \infty) \). So the correct option is A.

Step1: Recall zeros of a function

Zeros of a function \( g(x) \) are the values of \( x \) for which \( g(x) = 0 \). So we need to find \( x \) such that \( x^2 + 4x - 12 = 0 \).

Step2: Solve the quadratic equation

Factor the quadratic: \( x^2 + 4x - 12 = (x + 6)(x - 2) = 0 \). Setting each factor to zero: \( x + 6 = 0 \) gives \( x = -6 \), and \( x - 2 = 0 \) gives \( x = 2 \). Wait, but let's check the options. Wait, maybe I factored wrong? Wait, \( x^2 + 4x - 12 \): looking for two numbers that multiply to -12 and add to 4. Those numbers are 6 and -2. So \( (x + 6)(x - 2) = x^2 + 6x - 2x - 12 = x^2 + 4x - 12 \). So zeros at \( x = -6 \) and \( x = 2 \). Wait, but the options:

• Option A: \( g(0) = 2 \) and \( g(0) = -6 \): No, \( g(0) \) is a single value, and zeros are when \( g(x)=0 \), not \( g(0) \).
• Option B: \( g(0) = -2 \) and \( g(0) = 6 \): Same issue, \( g(0) \) is single, and zeros are \( x \) where \( g(x)=0 \), not \( g(0) \).
• Option C: \( g(-2) = 0 \) and \( g(6) = 0 \): Let's check \( g(-2) = (-2)^2 + 4(-2) - 12 = 4 - 8 - 12 = -16

eq 0 \). \( g(6) = 36 + 24 - 12 = 48
eq 0 \). So C is wrong.

• Option D: \( g(2) = 0 \) and \( g(-6) = 0 \): Check \( g(2) = 4 + 8 - 12 = 0 \). \( g(-6) = 36 - 24 - 12 = 0 \). So zeros are at \( x = 2 \) and \( x = -6 \), meaning \( g(2) = 0 \) and \( g(-6) = 0 \). So option D is correct.

Answer:

A. $(-\infty, \infty)$

Question 14