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

1 Answer
Mar 10, 2016

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

Tony BTony B