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

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Answer 1 · 2015-05-17T17:12:34

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

Re arrange the equation as follows

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

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

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

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

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

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

Answer 2 · 2015-05-17T17:54:54

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

$x_{v}$ $x_v$ $=$ $=$ $- \frac{b}{2 a}$ $-b / (2 a)$

$y_{v}$ $y_v$ $=$ $=$ $- Delta / (4a)$

(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 $[7/4, oo)$ .

What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola y=4x^2-2x+2y=4x2−2x+2y=4x^2-2x+2?