Hello!
First, let's analyze the graph: notice that the concavity is up, so the coefficient a is positive. We also can see that the roots of this graph are 2 and 4.
Other information: the minimum point is (3, -0.5).
Now, let's analyze each alternative:
a. y = 0.5(x+2)² +4
[tex]\begin{gathered} y=0.5\mleft(x+2\mright)^2+4 \\ y=0.5x^2+2x+6 \\ \end{gathered}[/tex]
Let's calculate the minimum point of it and compare:
[tex]\begin{gathered} x_V=\frac{-b}{2\cdot a}=-\frac{2}{2\cdot0.5}=\frac{-2}{1}_{}=-2 \\ \\ y_V=-\frac{\Delta}{4\cdot a}=-\frac{b^2-4\cdot a\cdot c}{4\cdot a}=-\frac{2^2-4\cdot0.5\cdot6}{4\cdot0.5}=-\frac{4-12}{2}=-\frac{-8}{2}=4 \end{gathered}[/tex]
Notice that these coordinates of the minimum point are different as (3, -0.5), so it's false.
b. y = 0.5(x+3)²-0.5
[tex]\begin{gathered} y=0.5\mleft(x+3\mright)^2-0.5 \\ y=0.5x^2+3x+4 \end{gathered}[/tex]
Let's calculate the minimum point too:
[tex]\begin{gathered} x_V=\frac{-b}{2\cdot a}=\frac{-3}{2\cdot0.5}=\frac{-3}{1}=-3 \\ \\ y_V=-\frac{\Delta}{4\cdot a}=-\frac{b^2-4\cdot a\cdot c}{4\cdot a}=-\frac{3^2-4\cdot0.5\cdot4}{4\cdot0.5}=-\frac{9-8}{2}=-\frac{1}{2} \end{gathered}[/tex]
Minimum point: (-3, -0.5) is different as (3, -0.5). False too.
c. y = 0.5(x-3)²-0.5
[tex]\begin{gathered} y=0.5\mleft(x-3\mright)^2-0.5 \\ y=0.5x^2-3x+4 \end{gathered}[/tex]
Calculating the minimum point:
[tex]\begin{gathered} x_V=\frac{-b}{2\cdot a}=-\frac{-3}{2\cdot0.5}=-\frac{-3}{1}=3 \\ \\ y_V=-\frac{\Delta}{4\cdot a}=-\frac{b^2-4\cdot a\cdot c}{4\cdot a}=-\frac{(-3)^2-4\cdot0.5\cdot4}{4\cdot0.5}=-\frac{9-8}{2}=-\frac{1}{2} \end{gathered}[/tex]
We obtained the point (3, -0.5) as the minimum point and it is equal, so, it's true.
Let me show you all these functions in the cartesian plane to confirm it:
Right answer: Alternative C.
