The quadratic equation is;
[tex]\begin{gathered} y\text{ = -}\frac{x^2}{2}\text{ - x + 5} \\ \\ \text{where a = -}\frac{1}{2}\text{ , b = -1 and c = 5} \end{gathered}[/tex]Here, given the information in the question, we want to write the quadratic equation in its standard form
To do this we shall make some substitutions;
The maximum point of the quadratic equation is also called the vertex
In this question, the vertex is g(-1) = 6
So this is (-1,6)
The other point is g(-3) = 4
So the point here is (-3,4)
Thus, we have a quadratic equation with vertex (-1,6) that passes through the point (-3,4)
The general equation for a quadratic equation having a vertex and passing through a given point is;
[tex]y=a(x-h)^2\text{ + k}[/tex]where (h,k) represents the coordinates of the vertex and (x,y) represents the coordinates of the points that the quadratic equation passes through
Thus, we have;
[tex]y=a(x+1)^2\text{ + 6}[/tex]so to get the a value in this form, we substitute the values of the points; where x is -3 and y is 4
We have;
[tex]\begin{gathered} 4=a(-3+1)^2\text{ + 6} \\ \\ 4\text{ = 4a + 6} \\ 4a\text{ = 4-6} \\ 4a\text{ = -2} \\ \\ a\text{ = }\frac{-2}{4} \\ \\ a\text{ = }\frac{-1}{2} \end{gathered}[/tex]So the quadratic equation can be written as;
[tex]y\text{ = }\frac{-1}{2}(x+1)^2\text{ + 6}[/tex]We can rewrite this in the form in the question as follows;
[tex]\begin{gathered} y\text{ = }\frac{-1}{2}(x+1)^2\text{ + 6} \\ \\ y\text{ = }\frac{-1(x^2\text{ + 2x + 2) }}{2}\text{ + 6} \\ \\ y\text{ = -}\frac{x^2}{2}\text{ - }\frac{2x}{2}\text{ - }\frac{2}{2}\text{ + 6} \\ \\ y\text{ = -}\frac{x^2}{2}\text{ - x -1 + 6} \\ \\ y\text{ = -}\frac{x^2}{2}\text{ - x + 5} \end{gathered}[/tex]