(Please show all work) 1. Find f(6) if f(x)= 1/2x + x - 4 2.Find the distance between (2,-4) and (5, -8) 3. What is the vertex of the parabola when the equation is y= (x+3)^2-1

Answer:[tex]\large\boxed{1.\ f(6)=5\ or\ f(6)=20}\\\boxed{2.\ d=5}\\\boxed{3.\ vertex=(-3,\ -1)}[/tex]Step-by-step explanation:[tex]1.\\\text{Put x = 6 to the equation of a function}\ f(x):\\\\\text{If}\ f(x)=\dfrac{1}{2}x+x-4\to f(6)=\dfrac{1}{2}(6)+6-4=3+6-4=5\\\\\text{If}\ f(x)=\dfrac{1}{2}x^2+x-4\to f(6)=\dfrac{1}{2}(6^2)+6-4=\dfrac{1}{2}(36)+2=18+2=20\\\\2.\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{We have the points (2, -4) and (5, -8). Substitute:}\\\\d=\sqrt{(5-2)^2+(-8-(-4))^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5[/tex][tex]3.\\\text{The vertex formula of a parabola:}\\\\y=a(x-h)^2+k\\\\(h,\ k)-vertex\\\\\text{We have}\ y=(x+3)^2-1=(x-(-3))^2-1\\\\\text{Therefore the vertex is:}\ (-3,\ -1).[/tex]