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

2 Answers
May 17, 2015

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

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 = -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) .