Question

given the function $p(x) = x^3 + 5x^2 - 14x$.
the $y$-intercept is $(0,-7)$ $(0,0)$ $(0,2)$ invalid notation.
the $x$-intercepts is/are
when $x \
ightarrow \infty, y \
ightarrow$ $\infty$
when $x \
ightarrow -\infty, y \
ightarrow$ $-\infty$
check answer

Explanation:

Step1: Find y - intercept

To find the y - intercept, we set \(x = 0\) in the function \(P(x)=x^{3}+5x^{2}-14x\).
Substitute \(x = 0\) into \(P(x)\): \(P(0)=0^{3}+5\times0^{2}-14\times0 = 0\). So the y - intercept is \((0,0)\).

Step2: Find x - intercepts

To find the x - intercepts, we set \(P(x)=0\), so we solve the equation \(x^{3}+5x^{2}-14x = 0\).
First, factor out an \(x\) from the left - hand side of the equation: \(x(x^{2}+5x - 14)=0\).
Then, factor the quadratic expression \(x^{2}+5x - 14\). We need two numbers that multiply to \(- 14\) and add up to \(5\). The numbers are \(7\) and \(-2\). So \(x^{2}+5x - 14=(x + 7)(x-2)\).
The factored form of the equation \(x^{3}+5x^{2}-14x = 0\) is \(x(x + 7)(x - 2)=0\).
Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we set each factor equal to zero:

• If \(x=0\), then the solution is \(x = 0\).
• If \(x + 7=0\), then \(x=-7\).
• If \(x - 2=0\), then \(x = 2\).

So the x - intercepts are the points where \(x=-7,0,2\) and \(y = 0\). In n - tuple (ordered triple for three x - intercepts) form, the x - intercepts are \((-7,0),(0,0),(2,0)\).

Step3: Analyze the end - behavior

For a polynomial function \(P(x)=a_{n}x^{n}+a_{n - 1}x^{n - 1}+\cdots+a_{1}x + a_{0}\), the end - behavior is determined by the leading term \(a_{n}x^{n}\) (the term with the highest power of \(x\)).
In the function \(P(x)=x^{3}+5x^{2}-14x\), the leading term is \(x^{3}\) with \(n = 3\) (odd) and \(a_{n}=1>0\).

• When \(x

ightarrow\infty\), for the leading term \(x^{3}\), as \(x\) gets very large positive, \(x^{3}\) gets very large positive. So \(y=x^{3}+5x^{2}-14x
ightarrow\infty\) as \(x
ightarrow\infty\).

• When \(x

ightarrow-\infty\), for the leading term \(x^{3}\), as \(x\) gets very large negative, \(x^{3}\) gets very large negative (because for an odd power, a negative number raised to an odd power is negative). So \(y=x^{3}+5x^{2}-14x
ightarrow-\infty\) as \(x
ightarrow-\infty\).

Answer:

s:

• The y - intercept is \(\boldsymbol{(0,0)}\).
• The x - intercepts are \(\boldsymbol{(-7,0),(0,0),(2,0)}\).
• When \(x

ightarrow\infty\), \(y
ightarrow\boldsymbol{\infty}\).

• When \(x

ightarrow-\infty\), \(y
ightarrow\boldsymbol{-\infty}\).