To solve this problem, first, we use the following form of the equation of a vertical parabola:
[tex]y=a(x-h)^2+k,[/tex]
where (h,k) are the coordinates of the vertex.
Using the above form and the graph, we get:
[tex]y=a(x+4)^2+2.[/tex]
Now, to determine the value of a, we consider the point (-2,0) on the parabola:
[tex]0=a(-2+4)^2+2.[/tex]
Solving for a, we get:
[tex]\begin{gathered} -2=a2^2, \\ -\frac{2}{4}=a, \\ a=-\frac{1}{2}. \end{gathered}[/tex]
Therefore:
[tex]y=-\frac{1}{2}(x+4)^2+2.[/tex]
Finally, we take the above equation to its standard form:
[tex]\begin{gathered} y=-\frac{1}{2}(x^2+8x+16)+2, \\ y=-\frac{1}{2}x^2-4x-8+2, \\ y=-\frac{1}{2}x^2-4x-6. \end{gathered}[/tex]
Answer:
[tex]y=-\frac{1}{2}x^2-4x-6.[/tex]
