What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola y=4x^2-2x+2?

May 17, 2015

Vertex $\left(\frac{1}{4} , \frac{7}{4}\right)$ Axis of symmetry x= $\frac{1}{4}$, Min 7/4, Max $\infty$

Re arrange the equation as follows

y= $4 \left({x}^{2} - \frac{x}{2}\right) + 2$

= $4 \left({x}^{2} - \frac{x}{2} + \frac{1}{16} - \frac{1}{16}\right)$ +2

=$4 \left({x}^{2} - \frac{x}{2} + \frac{1}{16}\right) - \frac{1}{4} + 2$

=$4 {\left(x - \frac{1}{4}\right)}^{2}$ +7/4

The vertex is $\left(\frac{1}{4} , \frac{7}{4}\right)$ Axis of symmetry is x=$\frac{1}{4}$

Minimum value is y=7/4 and maximum is $\infty$

May 17, 2015

In the general case, the coordinates of the vertex for a function of the 2nd degree $a {x}^{2} + b x + c$ are the following:

${x}_{v}$ $=$ $- \frac{b}{2 a}$

${y}_{v}$ $=$ $- \frac{\Delta}{4 a}$

(where $\Delta$ $=$ ${b}^{2} - 4 a c$)

In our particular case, the vertex will have the following coordinates:

${x}_{v}$ $=$ $- \frac{- 2}{2 \cdot 4}$ $=$ $\frac{1}{4}$

${y}_{v}$ $=$ $- \frac{{\left(- 2\right)}^{2} - 4 \cdot 4 \cdot 2}{4 \cdot 4}$ $=$ $\frac{7}{4}$

The vertex is the point $V \left(\frac{1}{4} , \frac{7}{4}\right)$

We can see that the function has a minimum, that is ${y}_{v}$ $=$ $\frac{7}{4}$

The axis of symmetry is a parallel line to the $O y$ axis passing through the vertex $V \left(\frac{1}{4.} \frac{7}{4}\right)$, i.e. the constant function $y$ $=$ $\frac{1}{4}$

As $y$ $\ge$ $\frac{7}{4}$, the range of our function is the interval $\left[\frac{7}{4} , \infty\right)$.