Question

How do you find the axis of symmetry, graph and find the maximum or minimum value of the function `f(x)=2x^2-8x+6`?

Answer 1

See below

Explanation:

`y=2x^2-8x+6`

How to draw a parabola:
1. Understand what kind of parabola is;
2. Find the coordinates of the vertex;
3. Draw the axis of symmetry;
4. Calculate two points that are distinct and opposite to the axis of symmetry;
You can continue with other points but that's enough.

Step 1:
The parabola is given with this kind of equation `y = ax^2+bx+c`, so we can affirm that the axis of symmetry is vertical. Moreover `a>0` hence the concavity is turned upwards

Step 2:
To find the vertex we can use this formula:
`V = (-\frac{b}{2a};-\frac{Delta}{4a})` where `Delta = b^2-4ac`

Let's find `Delta`
`Delta = (-8)^2-4(2)(6)`
`Delta = 64-48`
`Delta = 16`

Now the vertex:
`V_x = -\frac{-8}{2(2)} = \frac{8}{4} = 2`
`V_y = -\frac{16}{4(2)} = -\frac{16}{8} = -2`
`V = (2;-2)`

Right now we can jump to the conclusion that this is also the minimum point since the concavity is upward and even that there is no maximum point.
Infact the range of this parabola is `(-2;+oo)`

Step 3:
The axis of symmetry is with the x coordinate of the vertex `V_x` , so the equation of the line is:
`x = 2`

Step 4:
Let's find two points:
Make it simple with `x=0`

`y=2(0)^2-8(0)+6`
`y=6`
Point `A = (0,6)`

Another point with `x=4`

`y=2(4)^2-8(4)+6`
`y=2*16-32+6`
`y=\cancel{32}-\cancel{32}+6`
`y=6`
Point `B = (4,6)`

graph{y=2x^2-8x+6 [-4.1, 9.95, -2.64, 4.384]}