Question

How do you graph the parabola `y = 2(x - 1)^2 - 5` using vertex, intercepts and additional points?

Answer 1

See below.

Explanation:

The vertex form of a quadratic is given as:

`y=a(x-h)^2+k`

Where:

`bba` is the coefficient of `x^2`

`bbh` is the axis of symmetry.

`bbk` is the maximum/minimum value of the function.

The vertex of a parabola is the point where the curve turns, this is always at a maximum or a minimum value.

The coordinates of the vertex are:

`(h,k)`

Our function is in vertex form and from this we note that:

`h=1` and `k=-5`

So the coordinates of the vertex are:

`color(blue)((1,-5)`

Be careful when finding `h`. Remember that we have in the vertex form `(x-h)^2`, this means if we have `(x+2)^2`, then `h<0`, i.e. `h=-2`. `(x-(-2))^2=(x+2)^2`.

Other useful points for graphing are:

`y` axis intercept. This occurs where `x=0`:

`y=2((0)-1)^2-5`

`y=3`

Coordinates:

`color(blue)((0,3)`

`x` axis intercepts occur where `y=0`

`2(x-1)^2-5=0`

`(x-1)^2=5/2`

Taking roots:

`x-1=+-sqrt(5/2)`

`x=1+sqrt(5/2)=1+sqrt(10)/2`

`x=1-sqrt(5/2)=1-sqrt(10)/2`

Coordinates:

`color(blue)((1+sqrt(10)/2,0)` , `color(blue)((1-sqrt(10)/2,0)`

We know have four point in which to graph the function:

GRAPH: