Answer:
vertex of the function is [tex](-3,-3)[/tex]
Step-by-step explanation:
Given Function, f(x) = [tex]\frac{1}{2}x^2+3x+\frac{3}{2}[/tex]
Simplify the given function to find the vertex.
use completing the square method.
[tex]f(x)=\frac{1}{2}x^2+3x+\frac{3}{2}[/tex]
[tex]f(x)=\frac{1}{2}(x^2+6x+3)[/tex]
[tex]f(x)=\frac{1}{2}(x^2+6x+9-9+3)[/tex]
[tex]f(x)=\frac{1}{2}((x^2+6x+9)-9+3)[/tex]
[tex]f(x)=\frac{1}{2}((x+3)^2-6)[/tex]
[tex]f(x)=\frac{1}{2}(x+3)^2-3[/tex]
let [tex]y=\frac{1}{2}(x+3)^2-3[/tex]
[tex]2(y+3)=(x+3)^2[/tex]
[tex](x+3)^2=2(y+3)[/tex]
Therefore, vertex of the function is [tex](-3,-3)[/tex]
Answer:
(-3,-3) just took the test on edg
Step-by-step explanation:
