Question

(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 1

`\large\boxed{1.\ f(6)=5\ or\ f(6)=20}\\\boxed{2.\ d=5}\\\boxed{3.\ vertex=(-3,\ -1)}`

Explanation:

`1.\\\text{Put x = 6 to the equation of a function}\ f(x):\\\\\text{If}\ f(x)=(1)/(2)x+x-4\to f(6)=(1)/(2)(6)+6-4=3+6-4=5\\\\\text{If}\ f(x)=(1)/(2)x^2+x-4\to f(6)=(1)/(2)(6^2)+6-4=(1)/(2)(36)+2=18+2=20\\\\2.\\\text{The formula of a distance between two points:}\\\\d=√((x_2-x_1)^2+(y_2-y_1)^2)\\\\\text{We have the points (2, -4) and (5, -8). Substitute:}\\\\d=√((5-2)^2+(-8-(-4))^2)=√(3^2+(-4)^2)=√(9+16)=√(25)=5`

`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).`