QUESTION IMAGE
Question
consider the function (p(x) = (x - 1)^2(x - 3))
the (y)-intercept is the point
the (x)-intercept(s) is/are the point(s)
as (x \to infty), (y \to)
as (x \to -infty), (y \to)
question 11
given (f(x) = -9(x + 1)^4(x + 2)^4(2x - 1)(x + 4)^2)
list the zeroes (ex. 4, 6, -2):
degree:
end behavior:
as (x \to -infty), (f(x) \to)
as (x \to +infty), (f(x) \to)
⚡ Using what you learned: Graphing Polynomial Functions
Step 1: Analyze the first function $P(x)$
We are given:
Find the $y$-intercept
Set \( x = 0 \):
The $y$-intercept is the point \( (0, -3) \).
Find the $x$-intercept(s)
Set \( P(x) = 0 \):
This gives \( x = 1 \) and \( x = 3 \).
The $x$-intercepts are the points \( (1, 0), (3, 0) \).
Determine end behavior
The leading term of \( P(x) \) is found by multiplying the highest degree terms of each factor:
Since the degree is odd (\( 3 \)) and the leading coefficient is positive (\( 1 \)):
- As \( x \to \infty \), \( y \to \infty \)
- As \( x \to -\infty \), \( y \to -\infty \)
---
Step 2: Analyze the second function $f(x)$
We are given:
List the zeroes
Set each factor to zero:
- \( x + 1 = 0 \implies x = -1 \)
- \( x + 2 = 0 \implies x = -2 \)
- \( 2x - 1 = 0 \implies x = \frac{1}{2} \) (or \( 0.5 \))
- \( x + 4 = 0 \implies x = -4 \)
The zeroes are: \( -1, -2, 0.5, -4 \)
Find the degree
Sum the exponents of the variable factors:
Determine end behavior
The leading term is:
Since the degree is odd (\( 11 \)) and the leading coefficient is negative (\( -18 \)):
- As \( x \to -\infty \), \( f(x) \to \infty \)
- As \( x \to \infty \), \( f(x) \to -\infty \)
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For the first question:
- The $y$-intercept is the point:
(0, -3) - The $x$-intercept(s) is/are the point(s):
(1, 0), (3, 0) - As \( x \to \infty \), \( y \to \)
\infty(oroo) - As \( x \to -\infty \), \( y \to \)
-\infty(or-oo)
For Question 11:
- List the zeroes:
-1, -2, 0.5, -4 - Degree:
11 - End behavior:
- As \( x \to -\infty \), \( f(x) \to \)
\infty(oroo) - As \( x \to \infty \), \( f(x) \to \)
-\infty(or-oo)