Graph each polynomial function. Factor first if the expression is not in factored form.f(x)=x^{3} + 3x^{2} - 13x - 15 PSZ

The rational Root Theorem states that a rational root of a polynomial equation ${\displaystyle a_n{x}^n+a_{n-1}{x}^{n-1}+\dot{s}c+a_2{x}^2+a_{1}x+a_0{x}^0=0\text{with integer coeffcients is of the form}\frac{p}{q}\text{weher}p\text{is a factor of the constant term,}{a}_0,\text{and}q\text{is a factor of the leading coefficient,}{a}_n}$
${\displaystyle f\left(x\right)=x^3+3x^2-13x-15}$
${\displaystyle p\text{could equal any factors of}-15}$
${\displaystyle \text{So,}\pm 1,\pm 3,\pm 5,\pm 15}$ q could equal any foctors of 1 So, ${\displaystyle \pm 1}$ Therefore, possible rational zeros are ${\displaystyle \pm 1,\pm 3,\pm 5,\pm 15}$ Since the polynomial is ${\displaystyle x^3+3x^2-13x-15}$ so, the cooefficient are 1, 3, -13, -15 Let 1 is an actual zero of the polynomial and use synthetic division The remainder is 0 so, -1 is an actual zero and ${\displaystyle (x+1)\text{is a factor of}{x}^3+3x^2-13x-15}$ The equatient is ${\displaystyle x^2+2x-15}$
${\displaystyle f\left(x\right)=(x+1)(x^2+2x-15)}$
${\displaystyle x^2+2x-15=(x-3)(x+5)}$ Factor the trinomial So, ${\displaystyle f\left(x\right)=(x+1)(x-3)(x+5)}$ The function in factored form The function has three zeros -1, 3, and -5 So, the graph of f(x) crosses the x-axis at (-1, 0), (3, 0), and (-5, 0) To find the y-intercept, substitute 0 for x in f(x) ${\displaystyle f\left(x\right)=x^3+3x^2-13x-15}$
${\displaystyle f\left(0\right)=0^3+3{\left(0\right)}^2-13\left(0\right)-15}$ Substitute 0 for x ${\displaystyle =-15}$ So, the function f(x) crosses the y-axis at (0, -15) The leading coefficient is 1 Since the leading coefficient is positive and the function f(x) of degree 3 (odd degree) So, the end behavior is ${\displaystyle x\to \mathrm{\infty },f\left(x\right)\to \mathrm{\infty }}$
${\displaystyle x\to -\mathrm{\infty },f\left(x\right)\to -\mathrm{\infty }}$ See the graph below