the graph of the even function $f(x)$ has five $x$-intercepts. if $(6, 0)$ is one of the intercepts, which set of points can be the other $x$-intercepts of the graph of $f(x)$?
$(-4, 0)$, $(0, 0)$, and $(2, 0)$
$(-6, 0)$, $(-2, 0)$, and $(0, 0)$
$(-4, 0)$, $(-2, 0)$, and $(0, 0)$
$(-6, 0)$, $(-2, 0)$, and $(4, 0)$

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Accepted Answer

# Explanation:
## Step1: Recall the property of even functions
For an even function $ f(x) $, we have $ f(x)=f(-x) $ for all $ x $ in the domain. This implies that if $ (a, 0) $ is an $ x $-intercept (i.e., $ f(a) = 0 $), then $ (-a, 0) $ must also be an $ x $-intercept because $ f(-a)=f(a) = 0 $. Also, the graph of an even function is symmetric about the $ y $-axis.
Given that $ (6, 0) $ is an $ x $-intercept, by the property of even functions, $ (-6, 0) $ must also be an $ x $-intercept.

## Step2: Analyze the number of intercepts
The function has five $ x $-intercepts. We already have $ (6, 0) $ and $ (-6, 0) $ (from the even function property). Now we need three more intercepts. Among the options, we need to check which set includes $ (-6, 0) $ and has a total of three other intercepts (along with $ (6,0) $ and $ (-6,0) $ to make five).
Let's check each option:
- Option 1: $ (-4, 0),(0, 0),(2, 0) $: Does not include $ (-6, 0) $, so eliminate.
- Option 2: $ (-6, 0),(-2, 0),(0, 0) $: Includes $ (-6, 0) $. Now, the intercepts would be $ (6, 0), (-6, 0), (-2, 0), (0, 0) $ and we need one more? Wait, no, the set given is three points, so total intercepts: $ (6,0) $ (given), $ (-6,0), (-2,0), (0,0) $ from the set. Wait, no, the problem says "the other $ x $-intercepts" besides $ (6,0) $. So total intercepts: $ (6,0) $ plus the three in the set. So total $ 1 + 3=4 $? Wait, no, maybe I misread. Wait, the function has five $ x $-intercepts. One is $ (6,0) $, so we need four more? Wait, no, the options are sets of "the other $ x $-intercepts", so the number of intercepts in the set plus 1 (the $ (6,0) $) should be five. So the set should have four intercepts? Wait, no, let's re - examine.
Wait, the even function: if $ (a,0) $ is an intercept, $ (-a,0) $ is also an intercept, unless $ a = 0 $. The $ y $-intercept (when $ x = 0 $) is its own mirror image (since $ f(0)=f(-0)=f(0) $). So for non - zero $ a $, intercepts come in pairs $ (a,0) $ and $ (-a,0) $, and $ (0,0) $ is a single intercept.
So, let's consider the structure of the intercepts:
Case 1: If there is a non - zero intercept $ a
eq0 $, then $ -a $ is also an intercept.
We have $ (6,0) $, so $ (-6,0) $ is also an intercept (pair). Now, we need three more intercepts. If one of them is $ (0,0) $ (single), and two other non - zero intercepts which are their own pairs? No, non - zero intercepts must come in pairs. Wait, no, if we have $ (-2,0) $, then $ (2,0) $ should also be an intercept. But in option 2, the set is $ (-6,0), (-2,0),(0,0) $. Wait, but if $ (-2,0) $ is an intercept, $ (2,0) $ should also be an intercept (because of even function). But the option does not include $ (2,0) $. Wait, maybe the $ (-2,0) $ is such that $ 2 = - 2 $? No, $ 2
eq - 2 $ unless we consider $ x = 0 $. Wait, no, the mistake is in my earlier reasoning. Wait, the $ x $-intercept at $ x = 0 $ is symmetric about the $ y $-axis (since $ f(0)=f(-0) $). For non - zero $ x $, $ (a,0) $ and $ (-a,0) $ are both intercepts.
So, let's count the number of intercepts correctly:
- If we have $ (6,0) $, then $ (-6,0) $ (1 pair, 2 intercepts)
- If we have $ (-2,0) $, then $ (2,0) $ should be an intercept (another pair, 2 intercepts)
- And $ (0,0) $ (1 intercept)
But the function has five intercepts. So $ 2+2 + 1=5 $. But in option 2, the set is $ (-6,0), (-2,0),(0,0) $. So the intercepts would be $ (6,0), (-6,0), (-2,0), (2,0), (0,0) $. Wait, but the set given in option 2 does not include $ (2,0) $. Wait, maybe the option is considering that the "other" intercepts include $ (-6,0), (-2,0),(0,0) $, and we assume that $ (2,0) $ is implied? No, that's not right. Wait, maybe the $ (-2,0) $ is a typo or maybe I misinterpret. Wait, let's check the number of intercepts in each option:
- Option 2: The set has three points: $ (-6,0), (-2,0),(0,0) $. Along with $ (6,0) $, the total intercepts are $ (6,0), (-6,0), (-2,0), (0,0) $ – that's four intercepts. Wait, no, maybe the $ (-2,0) $ is such that $ x=-2 $ and since the function is even, $ x = 2 $ is also an intercept, but it's not listed. This is confusing. Wait, maybe the correct approach is:
The key is that for an even function, non - zero $ x $-intercepts come in pairs $ (a,0) $ and $ (-a,0) $. The $ x = 0 $ intercept is alone.
We have one intercept $ (6,0) $, so $ (-6,0) $ must be an intercept (pair). Now, we need three more intercepts. Among the options, the only option that includes $ (-6,0) $ is the second option ( $ (-6, 0),(-2, 0),(0, 0) $) and the fourth option ( $ (-6, 0),(-2, 0),(4, 0) $) is wrong because $ (4,0) $ should have $ (-4,0) $ as an intercept. The first option does not have $ (-6,0) $, the third option does not have $ (-6,0) $.
Wait, let's re - count the number of intercepts in each option when combined with $ (6,0) $:
- Option 1: Intercepts are $ (6,0), (-4,0),(0,0),(2,0) $ – 4 intercepts. Not five.
- Option 2: Intercepts are $ (6,0), (-6,0), (-2,0),(0,0) $ – wait, no, the set has three points, so $ (6,0) $ plus three points: $ (6,0), (-6,0), (-2,0),(0,0) $ – 4 intercepts. Wait, maybe the $ (-2,0) $ is a typo and should be $ (2,0) $, but no. Wait, maybe the function has a $ y $-intercept $ (0,0) $ (which is its own mirror image) and two pairs of non - zero intercepts. We have one pair $ (6,0) $ and $ (-6,0) $, and we need another pair and the $ y $-intercept. Wait, a pair is two intercepts, so one pair (2) + another pair (2) + $ (0,0) $ (1) = 5. So we need another pair, say $ (a,0) $ and $ (-a,0) $, and $ (0,0) $.
Looking at the options, the second option has $ (-6,0) $ (paired with $ (6,0) $), $ (-2,0) $ (which should be paired with $ (2,0) $, but it's not listed) and $ (0,0) $. But among the given options, this is the only one that includes $ (-6,0) $. The fourth option has $ (-6,0) $, $ (-2,0) $, $ (4,0) $ – $ (4,0) $ should have $ (-4,0) $ as an intercept, so it's missing. The first option has no $ (-6,0) $, the third option has no $ (-6,0) $.
Wait, maybe the correct answer is the second option: $ (-6, 0), (-2, 0), $ and $ (0, 0) $ because it includes $ (-6, 0) $ (the mirror of $ (6, 0) $) and has a $ y $-intercept $ (0, 0) $ and $ (-2, 0) $ (and its mirror $ (2, 0) $ which is implied by the even function, but maybe in the option, they just list $ (-2, 0) $ and we assume $ (2, 0) $ is also an intercept, but that's not correct. Wait, no, maybe the problem has a typo, but based on the property of even functions (needing $ (-6,0) $ as an intercept), the only option with $ (-6,0) $ is the second option ( $ (-6, 0), (-2, 0), (0, 0) $) and the fourth option ( $ (-6, 0), (-2, 0), (4, 0) $ is wrong because of $ (4,0) $). So the correct option is the second one: $ (-6, 0), (-2, 0), $ and $ (0, 0) $

# Answer: The set of points is $\boldsymbol{(-6, 0), (-2, 0), \text{and } (0, 0)}$ (the second option in the given choices)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the property of even functions

Step2: Analyze the number of intercepts

Answer: