Question

1. graph: function in standard form: ( q(x) = x^3 + 2x^2 + x + 2 ) function in factored form: ( (x + 2)(x^2 + 1) ) end behavior: as ( x \to -infty, q(x) \to -infty ) as ( x \to infty, q(x) \to infty ) roots (with multiplicity): ( x = i ), ( x = -2 ) degree: 3 y-intercept: 2

Explanation:

Step1: Analyze the leading term

The function is \( q(x)=x^{3}+2x^{2}+x + 2 \), the leading term is \( x^{3} \) with leading coefficient \( 1 \) (wait, but the given leading coefficient is -1? Wait, maybe there's a typo, but let's use the degree and leading coefficient for end - behavior. For a polynomial \( a_{n}x^{n}+a_{n - 1}x^{n-1}+\cdots+a_{0} \), the end - behavior is determined by \( a_{n} \) and \( n \). Here, degree \( n = 3 \) (odd) and if we consider the leading coefficient (let's assume the given leading coefficient - 1 is correct, maybe the standard form was miswritten). For a cubic function \( y=a x^{3}+bx^{2}+cx + d \), when \( a<0 \):

• As \( x

ightarrow\infty \), \( y
ightarrow-\infty \) (because \( x^{3} \) dominates, and \( a=- 1<0 \), so \( (-1)\times\infty^{3}=-\infty \))

• As \( x

ightarrow-\infty \), \( y
ightarrow\infty \) (because \( (-1)\times(-\infty)^{3}=(-1)\times(-\infty)=\infty \))

Wait, but the original function's standard form is \( x^{3}+2x^{2}+x + 2 \), leading coefficient \( 1 \). Let's recast: For \( y = x^{3}+2x^{2}+x + 2 \), degree \( 3 \) (odd), leading coefficient \( 1>0 \).

• As \( x

ightarrow\infty \), \( y=\infty^{3}+2\infty^{2}+\infty + 2
ightarrow\infty \)

• As \( x

ightarrow-\infty \), \( y=(-\infty)^{3}+2(-\infty)^{2}+(-\infty)+2=-\infty + 2\infty-\infty + 2=(-\infty-\infty)+(2\infty + 2)=-2\infty+2\infty + 2
ightarrow-\infty \) (wait, no: \( (- \infty)^{3}=-\infty \), \( 2(-\infty)^{2}=2\infty \), \( (-\infty) \), so \( -\infty+2\infty-\infty + 2=(2\infty)+(-\infty-\infty)+2=2\infty-2\infty + 2 = 2 \)? No, that's wrong. The correct way: for large \( |x| \), the \( x^{3} \) term dominates. So for \( y=x^{3} \) (when \( x
ightarrow\infty \), \( y
ightarrow\infty \); when \( x
ightarrow-\infty \), \( y
ightarrow-\infty \)).

But the problem states the leading coefficient is -1. Let's use the given leading coefficient -1 and degree 3.

• For a cubic function \( y=-x^{3}+bx^{2}+cx + d \) (degree 3, odd, \( a=-1<0 \)):
• As \( x

ightarrow\infty \), \( y=-(\infty)^{3}=-\infty \)

• As \( x

ightarrow-\infty \), \( y=-(-\infty)^{3}=\infty \)

Step2: Find the roots

We can factor the function \( q(x)=x^{3}+2x^{2}+x + 2 \) by grouping:
\( q(x)=x^{2}(x + 2)+1(x + 2)=(x^{2}+1)(x + 2) \)
Setting \( q(x)=0 \), we have \( x^{2}+1 = 0\) or \( x + 2=0 \)

• For \( x^{2}+1=0 \), \( x^{2}=-1\), so \( x = i \) or \( x=-i \)
• For \( x + 2=0 \), \( x=-2 \)

So the roots are \( x = i \) (multiplicity 1), \( x=-i \) (multiplicity 1), \( x=-2 \) (multiplicity 1)

Step3: Find the y - intercept

To find the y - intercept, set \( x = 0 \) in \( q(x) \)
\( q(0)=0^{3}+2\times0^{2}+0 + 2=2 \)

Step4: Graph the function

• The roots are \( x=-2 \) (real root), \( x = i \) and \( x=-i \) (complex roots). So the graph crosses the x - axis at \( x=-2 \) (since multiplicity 1, it crosses the axis here). The y - intercept is at \( (0,2) \). The end - behavior: since degree 3 and leading coefficient 1 (from the standard form \( x^{3}+2x^{2}+x + 2 \)), as \( x

ightarrow\infty \), \( q(x)
ightarrow\infty \); as \( x
ightarrow-\infty \), \( q(x)
ightarrow-\infty \). We can plot the real root \( x=-2 \), the y - intercept \( (0,2) \), and use the end - behavior to sketch the graph. When \( x=-2 \), \( q(-2)=(-2)^{3}+2\times(-2)^{2}+(-2)+2=-8 + 8-2 + 2=0 \). For \( x=-3 \), \( q(-3)=(-3)^{3}+2\times(-3)^{2}+(-3)+2=-27 + 18-3 + 2=-10 \). For \( x = 1 \), \( q(1)=1 + 2+1 + 2=6 \).

Answer:

• End behavior: As \( x

ightarrow\infty \), \( q(x)
ightarrow\infty \); As \( x
ightarrow-\infty \), \( q(x)
ightarrow-\infty \) (if leading coefficient is 1) or As \( x
ightarrow\infty \), \( q(x)
ightarrow-\infty \); As \( x
ightarrow-\infty \), \( q(x)
ightarrow\infty \) (if leading coefficient is - 1)

• Factored form: \( q(x)=(x^{2}+1)(x + 2) \)
• Roots: \( x=-2 \) (multiplicity 1), \( x = i \) (multiplicity 1), \( x=-i \) (multiplicity 1)
• Degree: 3
• y - intercept: 2