What are the vertex and x-intercepts of the graph of y=(x-4)(x+2)?A. X-intercepts: (4,0),(-2,0)B. Vertex: (-1,-5)C. Vertex: (1,-9)D. Vertex: (1,9)E. x-intercepts: (-4,0)(-2,0)F. X-intercepts: (-4,0),(2,0)

Answer:A. x-intercepts: (4,0),(-2,0)C. Vertex: (1,-9)Step-by-step explanation:We have been given factored form of a quadratic equation [tex]y=(x-4)(x+2)[/tex]. We are asked to find vertex and x-intercepts of our given equation.We know that x-intercepts of a graph are those points at which graph touches or crosses x-axis and y-coordinates of x-intercepts are always zero.So we will substitute [tex]y=0[/tex] in our given equation to find the x-intercepts.[tex]0=(x-4)(x+2)[/tex][tex](x-4)=0\text{ or }(x+2)=0[/tex][tex]x-4+4=0+4\text{ or }x+2-2=0-2[/tex][tex]x=4\text{ or }x=-2[/tex]Since y-coordinates of x-intercepts will be 0, therefore, x-intercepts of the graph will be (4,0), (-2,0) and option A is the correct choice.To find the vertex of the parabola, we will expand our given expression using FOIL.[tex](x-4)(x+2)[/tex][tex]x*x+x*2-4*x-4*2[/tex][tex]x^2+2x-4x-8[/tex][tex]x^2-2x-8[/tex]We will use formula [tex]\frac{-b}{2a}[/tex] to find the x-coordinate of vertex of parabola, where, a and b represents the coefficient of [tex]x^2\text{ and }x[/tex] respectively.[tex]\text{x-coordinate of vertex}=\frac{-(-2)}{2*1}[/tex][tex]\text{x-coordinate of vertex}=\frac{2}{2}=1[/tex]Now we will substitute [tex]x=1[/tex] in the expression to find the y-coordinate of parabola.[tex]\text{y-coordinate of vertex}=1^2-2*1-8[/tex][tex]\text{y-coordinate of vertex}=1-2-8[/tex][tex]\text{y-coordinate of vertex}=-9[/tex]Therefore, the vertex of parabola will be at point (1,-9) and option C is the correct choice.