Answer :
We want to graph the cubic function
[tex]$$
f(x)=x^3+9x^2+2x-48.
$$[/tex]
A good starting point is to factor the polynomial. It turns out that the function factors as
[tex]$$
f(x)=(x-2)(x+3)(x+8).
$$[/tex]
This factorization tells us that the roots (or the [tex]\(x\)[/tex]-intercepts) occur when any of the factors is zero. Setting each factor equal to zero:
1. [tex]\( x-2=0 \)[/tex] gives [tex]\( x=2 \)[/tex],
2. [tex]\( x+3=0 \)[/tex] gives [tex]\( x=-3 \)[/tex],
3. [tex]\( x+8=0 \)[/tex] gives [tex]\( x=-8 \)[/tex].
Thus, the [tex]\(x\)[/tex]-intercepts of the graph are at the points [tex]\((-8, 0)\)[/tex], [tex]\((-3, 0)\)[/tex], and [tex]\((2, 0)\)[/tex].
Next, to find the [tex]\(y\)[/tex]-intercept, we substitute [tex]\(x=0\)[/tex] into the function:
[tex]$$
f(0) = 0^3 + 9(0)^2 + 2(0) - 48 = -48.
$$[/tex]
Therefore, the [tex]\(y\)[/tex]-intercept is at the point [tex]\((0, -48)\)[/tex].
Now considering the overall behavior:
- Since this is a cubic function, the graph is a smooth curve.
- The leading coefficient (of [tex]\(x^3\)[/tex]) is positive; hence, as [tex]\(x\to\infty\)[/tex], [tex]\(f(x)\to\infty\)[/tex] and as [tex]\(x\to-\infty\)[/tex], [tex]\(f(x)\to-\infty\)[/tex].
In summary, the graph of the function [tex]\( f(x)=x^3+9x^2+2x-48 \)[/tex] has the following key features:
[tex]$$
\textbf{Factored form: } f(x)=(x-2)(x+3)(x+8)
$$[/tex]
[tex]$$
\textbf{\(x\)-intercepts: } x=-8,\; -3,\; 2 \quad \text{(points: } (-8,0),\, (-3,0),\, (2,0) \text{)}
$$[/tex]
[tex]$$
\textbf{\(y\)-intercept: } (0,-48).
$$[/tex]
When you graph this function, you should see a smooth curve that crosses the [tex]\(x\)[/tex]-axis at [tex]\(x=-8\)[/tex], [tex]\(x=-3\)[/tex], and [tex]\(x=2\)[/tex], and crosses the [tex]\(y\)[/tex]-axis at [tex]\((0,-48)\)[/tex]. The graph will start from the lower left (approaching [tex]\(-\infty\)[/tex] as [tex]\(x\to -\infty\)[/tex]) and rise to the upper right (approaching [tex]\(+\infty\)[/tex] as [tex]\(x\to \infty\)[/tex]).
[tex]$$
f(x)=x^3+9x^2+2x-48.
$$[/tex]
A good starting point is to factor the polynomial. It turns out that the function factors as
[tex]$$
f(x)=(x-2)(x+3)(x+8).
$$[/tex]
This factorization tells us that the roots (or the [tex]\(x\)[/tex]-intercepts) occur when any of the factors is zero. Setting each factor equal to zero:
1. [tex]\( x-2=0 \)[/tex] gives [tex]\( x=2 \)[/tex],
2. [tex]\( x+3=0 \)[/tex] gives [tex]\( x=-3 \)[/tex],
3. [tex]\( x+8=0 \)[/tex] gives [tex]\( x=-8 \)[/tex].
Thus, the [tex]\(x\)[/tex]-intercepts of the graph are at the points [tex]\((-8, 0)\)[/tex], [tex]\((-3, 0)\)[/tex], and [tex]\((2, 0)\)[/tex].
Next, to find the [tex]\(y\)[/tex]-intercept, we substitute [tex]\(x=0\)[/tex] into the function:
[tex]$$
f(0) = 0^3 + 9(0)^2 + 2(0) - 48 = -48.
$$[/tex]
Therefore, the [tex]\(y\)[/tex]-intercept is at the point [tex]\((0, -48)\)[/tex].
Now considering the overall behavior:
- Since this is a cubic function, the graph is a smooth curve.
- The leading coefficient (of [tex]\(x^3\)[/tex]) is positive; hence, as [tex]\(x\to\infty\)[/tex], [tex]\(f(x)\to\infty\)[/tex] and as [tex]\(x\to-\infty\)[/tex], [tex]\(f(x)\to-\infty\)[/tex].
In summary, the graph of the function [tex]\( f(x)=x^3+9x^2+2x-48 \)[/tex] has the following key features:
[tex]$$
\textbf{Factored form: } f(x)=(x-2)(x+3)(x+8)
$$[/tex]
[tex]$$
\textbf{\(x\)-intercepts: } x=-8,\; -3,\; 2 \quad \text{(points: } (-8,0),\, (-3,0),\, (2,0) \text{)}
$$[/tex]
[tex]$$
\textbf{\(y\)-intercept: } (0,-48).
$$[/tex]
When you graph this function, you should see a smooth curve that crosses the [tex]\(x\)[/tex]-axis at [tex]\(x=-8\)[/tex], [tex]\(x=-3\)[/tex], and [tex]\(x=2\)[/tex], and crosses the [tex]\(y\)[/tex]-axis at [tex]\((0,-48)\)[/tex]. The graph will start from the lower left (approaching [tex]\(-\infty\)[/tex] as [tex]\(x\to -\infty\)[/tex]) and rise to the upper right (approaching [tex]\(+\infty\)[/tex] as [tex]\(x\to \infty\)[/tex]).
