Question

How do you use the important points to sketch the graph of `y=x^2+2`?

Answer 1

See explanantion

Explanation:

Determining the key points improves the precision for your sketch. It also demonstrates that you have some grasp of what the equation is doing.

Standard form;`" "-> ax^2+bx+c`

`color(blue)("Point 1")`
The coefficient of `x` is +1 (positive). This means that the graph of shape type `uu`

`color(blue)("Point 2")`
There is not any `bx` term. This means that the axis of symmetry is the y-axis. If there had been a `bx` term then the axis of symmetry would have been `(-1/2)xxb/a`

`color(blue)("Point 3")`
The constant lifts or lowers the whole graph depending on its sign
As we have a constant of + 2 it lifts the whole graph up by 2

`color(blue)("Point 4")`
As the graph is of form `y=ax^2+c` and it has been lifted up by 2
it has no x-axis intercept.

`color(blue)("Point 5")`
The y-intercept is at `x=0" "->" "y=(0)^2+2`.
y-intercept = 2

`color(blue)("Point 6")`
As you only have one key point (y-intercept) I would suggest that you do a quick determination of two other points to guide your hand drawn sketch. For example: `x=2" then you have "y=+-(2)^2+2 =+-6`