Let's analyze the function [tex]\( P(x) = x^3 + x^2 - 42x \)[/tex].
1. Finding the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept of a function is the value of the function when [tex]\( x = 0 \)[/tex].
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
P(0) = 0^3 + 0^2 - 42 \cdot 0 = 0
\][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( 0 \)[/tex].
2. Finding the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts are the values of [tex]\( x \)[/tex] where the function [tex]\( P(x) = 0 \)[/tex].
We solve the equation:
[tex]\[
x^3 + x^2 - 42x = 0
\][/tex]
Factor out the common term [tex]\( x \)[/tex]:
[tex]\[
x(x^2 + x - 42) = 0
\][/tex]
This gives us one [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex].
Now, we solve the quadratic equation [tex]\( x^2 + x - 42 \)[/tex]:
[tex]\[
x^2 + x - 42 = 0
\][/tex]
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -42 \)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-42)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 168}}{2} = \frac{-1 \pm \sqrt{169}}{2} = \frac{-1 \pm 13}{2}
\][/tex]
Therefore, we get:
[tex]\[
x = \frac{-1 + 13}{2} = 6 \quad \text{and} \quad x = \frac{-1 - 13}{2} = -7
\][/tex]
So, the [tex]\( x \)[/tex]-intercepts are [tex]\( -7 \)[/tex], [tex]\( 0 \)[/tex], and [tex]\( 6 \)[/tex].
3. Analyzing the end behavior as [tex]\( x \rightarrow \infty \)[/tex]:
As [tex]\( x \)[/tex] approaches infinity, the term [tex]\( x^3 \)[/tex] will dominate because it has the highest power.
Therefore:
[tex]\[
\lim_ {x \to \infty} (x^3 + x^2 - 42x) = \infty
\][/tex]
4. Analyzing the end behavior as [tex]\( x \rightarrow -\infty \)[/tex]:
Similarly, as [tex]\( x \)[/tex] approaches negative infinity, the term [tex]\( x^3 \)[/tex] will still dominate.
Therefore:
[tex]\[
\lim_ {x \to -\infty} (x^3 + x^2 - 42x) = -\infty
\][/tex]
Summarizing all the findings:
- The [tex]\( y \)[/tex]-intercept is [tex]\( \boxed{0} \)[/tex].
- The [tex]\( x \)[/tex]-intercepts are [tex]\( \boxed{-7, 0, 6} \)[/tex].
- When [tex]\( x \rightarrow \infty \)[/tex], [tex]\( y \rightarrow \boxed{\infty} \)[/tex].
- When [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( y \rightarrow \boxed{-\infty} \)[/tex].
