Answer:
[tex]y=-\dfrac{1}{2}(x-1)^2+2[/tex]
Step-by-step explanation:
Vertex form of a quadratic relation
[tex]y=a(x-h)^2+k[/tex]
where:
- (h, k) is the vertex
- a is some constant
a > 0 : parabola opens upwards
a < 0 : parabola opens downwards
From inspection of the graph:
- Vertex = (1, 2)
- Point on the curve = (-3, -6)
Substitute the given vertex and point into the vertex formula to find a:
[tex]\implies -6=a(-3-1)^2+2[/tex]
[tex]\implies -6=16a+2[/tex]
[tex]\implies 16a=-8[/tex]
[tex]\implies a=-\dfrac{1}{2}[/tex]
Substitute the found value of a and the given vertex into the formula to create the equation for the graph in vertex form:
[tex]y=-\dfrac{1}{2}(x-1)^2+2[/tex]
