Question

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

Answer 1

Vertex `(1/4, 7/4)` Axis of symmetry x= `1/4`, Min 7/4, Max `oo`

Re arrange the equation as follows

y= `4(x^2 -x/2) +2`

= `4(x^2-x/2 +1/16-1/16)` +2

=`4(x^2 -x/2 +1/16)-1/4+2`

=`4(x-1/4)^2` +7/4

The vertex is `(1/4,7/4)` Axis of symmetry is x=`1/4`

Minimum value is y=7/4 and maximum is `oo`

Answer 2

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` `=` `-b / (2 a)`

`y_v` `=` `- Delta / (4a)`

(where `Delta` `=` `b^2 - 4 a c`)

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

`x_v` `=` `- (-2) / (2 * 4)` `=` `1 / 4`

`y_v` `=` `- ((-2)^2 - 4 * 4 * 2) / (4 * 4)` `=` `7 / 4`

The vertex is the point `V (1/4, 7/4)`

We can see that the function has a minimum, that is `y_v` `=` `7 / 4`

The axis of symmetry is a parallel line to the `Oy` axis passing through the vertex `V (1/4. 7/4)`, i.e. the constant function `y` `=` `1/4`

As `y` `>=` `7/4`, the range of our function is the interval `[7/4, oo)`.