Question

How do you find the axis of symmetry, graph and find the maximum or minimum value of the function `y = 2x^2 - 4x -3`?

Answer 1

Axis of symmetry`color(blue)(" "x=1)`
Minimum value of the function `color(blue)(=-5)`
See the explanation for the graph

Explanation:

The solution:

To find the Axis of symmetry you need to solve for the Vertex `(h, k)`

Formula for the vertex:
`h=(-b)/(2a)` and `k=c-b^2/(4a)`

From the given `y=2x^2-4x-3`
`a=2` and `b=-4` and `c=-3`

`h=(-b)/(2a)=(-(-4))/(2(2))=1`
`k=c-b^2/(4a)=-3-(-4)^2/(4(2))=-5`

Axis of symmetry:

`x=h`

`color(blue)(x=1)`

Since `a` is positive, the function has a Minimum value and does not have a Maximum.

Minimum value `color(blue)(=k=-5)`

The graph of `y=2x^2-4x-3`

To draw the graph of `y=2x^2-4x-3`, use the vertex `(h, k)=(1, -5)` and the intercepts.

When `x=0`,
`y=2x^2-4x-3`
`y=2(0)^2-4(0)-3=-3" "`means there is a point at `(0, -3)`

and when `y=0`,
`y=2x^2-4x-3`
`0=2x^2-4x-3`

`x=(-b+-sqrt(b^2-4ac))/(2a)=(-(-4)+-sqrt((-4)^2-4(2)(-3)))/(2(2))`

`x=(+4+-sqrt(16+24))/(4)`

`x=(+4+-sqrt(40))/(4)`

`x=(+4+-2sqrt(10))/(4)`

`x_1=1+1/2sqrt(10)`

`x_2=1-1/2sqrt(10)`

We have two points at `(1+1/2sqrt(10), 0)` and `(1-1/2sqrt(10), 0)`

God bless...I hope the explanation is useful.