Explanation
We need to find the zeros of the following function:
[tex]g(x)=-x^3+x^2+12x[/tex]
First of all it's important to note that all the terms have a power of x i.e. there's no constant term. This means that x=0 is a zero of the function:
[tex]g(0)=-0^3+0^2+12\cdot0=0+0+0=0[/tex]
So we have our first zero. Then you can notice that x is a common factor to all terms so we can rewrite our function:
[tex]\begin{gathered} g(x)=-x^3+x^2+12x=-x^2\cdot x+x\cdot x+12\cdot x=(-x^2+x+12)\cdot x \\ g(x)=(-x^2+x+12)\cdot x \end{gathered}[/tex]
So we can write g(x) as a product of two expressions. This means that if one of the expression is 0 for a given x then g(x) is also 0 for that x. Then the zeros of each term are also the zeros of g(x). The left term is just x and its zero is x=0 the one that we already found. Then the remaining zeros of g(x) are the x-values that make the quadratic expression on the left equal to 0:
[tex]-x^2+x+12=0[/tex]
Remember that for a quadratic equation ax²+bx+c=0 its zeros are given by the following formula:
[tex]r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case we have a=-1, b=1 and c=12 so the zeros of the quadratic expression are:
[tex]r=\frac{-1\pm\sqrt{1^2-4\cdot(-1)\cdot12}}{2\cdot(-1)}=\frac{-1\pm\sqrt{49}}{-2}=\frac{-1\pm7}{-2}[/tex]
Then we have another two zeros:
[tex]\begin{gathered} r=\frac{-1+7}{-2}=\frac{6}{-2}=-3 \\ r=\frac{-1-7}{-2}=-\frac{8}{-2}=4 \end{gathered}[/tex]
Then the three zeros of g(x) are x=0, x=-3 and x=4.
In a graph the zeros are the x-values of the points where the graph intercepts the x-axis so the answer is the function that intercepts the x-axis at -3, 0 and 4.
Answer
Then the answer is option A.
