Question

For the following quadratic function, (a) find the vertex, the axis of symmetry, and the maximum or minimum function value, and (b) graph the function.

h(x) =1/2x^2-2x+13/2

Answer 1

a)

`\displaystyle x_0=-(b)/(2a) =(2)/(2*(1)/(2) ) =2\\\\y_0=(1)/(2) *2^2-2*2+(13)/(2) =(9)/(2) \\\\\text{vertex:}\: (2;(9)/(2) )= > \text{axis of symmetry:}\:x=2`

`y_(min)=(9)/(2) (\text{because }a=(1)/(2) > 0)\\\\y_(max)=\infty`

b) on photo