Math / Algebra

QuestionWhich of the following could have zeros different from $y=p(x)$ for $k \neq 1$ ? $k p(x)$ $p(x-k)$ $p(x)+k$ $p(k x)$ None of these

Studdy Solution

STEP 1

What is this asking? We're trying to find out which of these functions could have different zeros than $y = p(x)$ when $k \neq 1$ . Watch out! Don't assume $k = 1$ or mix up transformations like shifting and scaling!

STEP 2

1. Analyze $k \cdot p(x)$
2. Analyze $p(x-k)$
3. Analyze $p(x) + k$
4. Analyze $p(k \cdot x)$

STEP 3

Let's start with $k \cdot p(x)$ .
The zeros of a function are where it equals zero.
So, we have:
$k \cdot p(x) = 0$
Now, if $k \neq 0$ , we divide both sides by $k$ (because dividing by a non-zero number doesn't change the zero locations):
$p(x) = 0$
This means the zeros of $k \cdot p(x)$ are the same as $p(x)$ .
So, no change in zeros here!

STEP 4

Now, let's look at $p(x-k)$ .
The zeros are where:
$p(x-k) = 0$
This means $x-k$ must be a zero of $p$ .
If $x_0$ is a zero of $p$ , then:
$x-k = x_0 \quad \Rightarrow \quad x = x_0 + k$
Aha! The zeros are shifted by $k$ .
So, different zeros when $k \neq 0$ !

STEP 5

Next, consider $p(x) + k$ .
The zeros occur where:
$p(x) + k = 0 \quad \Rightarrow \quad p(x) = -k$
This is different from $p(x) = 0$ .
So, unless $k = 0$ , the zeros are different!

STEP 6

Finally, let's check out $p(k \cdot x)$ .
The zeros are where:
$p(k \cdot x) = 0$
If $x_0$ is a zero of $p$ , then:
$k \cdot x = x_0 \quad \Rightarrow \quad x = \frac{x_0}{k}$
This means the zeros are scaled by $\frac{1}{k}$ .
So, different zeros when $k \neq 1$ !

STEP 7

The functions $p(x-k)$ , $p(x) + k$ , and $p(k \cdot x)$ could have zeros different from $y = p(x)$ when $k \neq 1$ .

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QuestionWhich of the following could have zeros different from $y=p(x)$ for $k \neq 1$ ? $k p(x)$ $p(x-k)$ $p(x)+k$ $p(k x)$ None of these

Studdy Solution

STEP 1

What is this asking? We're trying to find out which of these functions could have different zeros than $y = p(x)$ when $k \neq 1$ . Watch out! Don't assume $k = 1$ or mix up transformations like shifting and scaling!

STEP 2

1. Analyze $k \cdot p(x)$
2. Analyze $p(x-k)$
3. Analyze $p(x) + k$
4. Analyze $p(k \cdot x)$

STEP 3

STEP 4

Now, let's look at $p(x-k)$ .
The zeros are where:
$p(x-k) = 0$
This means $x-k$ must be a zero of $p$ .
If $x_0$ is a zero of $p$ , then:
$x-k = x_0 \quad \Rightarrow \quad x = x_0 + k$
Aha! The zeros are shifted by $k$ .
So, different zeros when $k \neq 0$ !

STEP 5

Next, consider $p(x) + k$ .
The zeros occur where:
$p(x) + k = 0 \quad \Rightarrow \quad p(x) = -k$
This is different from $p(x) = 0$ .
So, unless $k = 0$ , the zeros are different!

STEP 6

Finally, let's check out $p(k \cdot x)$ .
The zeros are where:
$p(k \cdot x) = 0$
If $x_0$ is a zero of $p$ , then:
$k \cdot x = x_0 \quad \Rightarrow \quad x = \frac{x_0}{k}$
This means the zeros are scaled by $\frac{1}{k}$ .
So, different zeros when $k \neq 1$ !

STEP 7

The functions $p(x-k)$ , $p(x) + k$ , and $p(k \cdot x)$ could have zeros different from $y = p(x)$ when $k \neq 1$ .

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QuestionWhich of the following could have zeros different from y=p(x)y=p(x)y=p(x) for k≠1k \neq 1k=1 ? kp(x)k p(x)kp(x) p(x−k)p(x-k)p(x−k) p(x)+kp(x)+kp(x)+k p(kx)p(k x)p(kx) None of these

Studdy Solution

STEP 1

What is this asking? We're trying to find out which of these functions could have different zeros than y=p(x) y = p(x) y=p(x) when k≠1 k \neq 1 k=1. Watch out! Don't assume k=1 k = 1 k=1 or mix up transformations like shifting and scaling!

STEP 2

1. Analyze k⋅p(x) k \cdot p(x) k⋅p(x)2. Analyze p(x−k) p(x-k) p(x−k)3. Analyze p(x)+k p(x) + k p(x)+k4. Analyze p(k⋅x) p(k \cdot x) p(k⋅x)

STEP 3

STEP 4

STEP 5

Next, consider p(x)+k p(x) + k p(x)+k.The zeros occur where:p(x)+k=0⇒p(x)=−kp(x) + k = 0 \quad \Rightarrow \quad p(x) = -kp(x)+k=0⇒p(x)=−kThis is different from p(x)=0 p(x) = 0 p(x)=0.So, unless k=0 k = 0 k=0, the zeros are **different**!

STEP 6

STEP 7

The functions p(x−k) p(x-k) p(x−k), p(x)+k p(x) + k p(x)+k, and p(k⋅x) p(k \cdot x) p(k⋅x) could have zeros different from y=p(x) y = p(x) y=p(x) when k≠1 k \neq 1 k=1.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionWhich of the following could have zeros different from $y=p(x)$ for $k \neq 1$ ? $k p(x)$ $p(x-k)$ $p(x)+k$ $p(k x)$ None of these

What is this asking? We're trying to find out which of these functions could have different zeros than $y = p(x)$ when $k \neq 1$ . Watch out! Don't assume $k = 1$ or mix up transformations like shifting and scaling!

1. Analyze $k \cdot p(x)$
2. Analyze $p(x-k)$
3. Analyze $p(x) + k$
4. Analyze $p(k \cdot x)$

Next, consider $p(x) + k$ .
The zeros occur where:
$p(x) + k = 0 \quad \Rightarrow \quad p(x) = -k$
This is different from $p(x) = 0$ .
So, unless $k = 0$ , the zeros are different!

The functions $p(x-k)$ , $p(x) + k$ , and $p(k \cdot x)$ could have zeros different from $y = p(x)$ when $k \neq 1$ .