write the equation of the quadratic equation function i standard form represented by the graph

Accepted Solution

Answer:[tex]x^{2} +6x+8=0[/tex]Step-by-step explanation:we know thatThe quadratic equation in standard form is equal to[tex]ax^{2} +bx+c=0[/tex]In this problem we havethe vertex is equal to the point [tex](-3,-1)[/tex]The zeros of the function are the points [tex](-4,0)[/tex] and [tex](-2,0)[/tex]The graph of the figure is a vertical parabola open upwardthe equation of a vertical parabola in vertex form is equal to[tex]y=a(x-h)^{2}+k[/tex]where (h,k) is the vertex of the parabolasubstitute[tex]y=a(x+3)^{2}-1[/tex]with the point [tex](-2,0)[/tex] find the value of a[tex]0=a(-2+3)^{2}-1[/tex][tex]0=a(1)^{2}-1[/tex][tex]a=1[/tex]the equation in vertex form is equal to[tex]y=(x+3)^{2}-1[/tex]Convert to standard form[tex]y=x^{2} +6x+9-1[/tex]equate to zero[tex]x^{2} +6x+9-1=0[/tex][tex]x^{2} +6x+8=0[/tex]