Chapter 3 - Quadratic Functions - Review Exercises and Problems for Chapter Three - Page 130: 23

i) Vertex: $(0.6,0)$ ii) Axis of symmetry: $x= 0.6$ iii) $x= 0$, $y$-intercept: $y=(0.6)^2=0.36$ iv) Concave up.

Give $y=(x-0.6)^2$ i) Vertex: $(0.6,0)$ ii) Axis of symmetry: $x= 0.6$ iii)$x= 0$, $y$-intercept: $y=(0.6)^2=0.36$ iv) Since the coefficient $a$ is positive the curve is concave up.