Question

How do you use the important points to sketch the graph of `f(x) = 12x^2 + 2`?

Answer 1

If I were you, I'd use the y-intercept and two points on the negative and positive parts of the graphs.

Explanation:

First, I'd substitute `x` with `0` to find the y-intercept. The answer is 12. Your first point will be `(0,12)`

Then, I'd pick two easy numbers on each side to graph. For example, I'd use `2` & `4` and `-2` & `-4`.

Now, I'd substitute the `x` with 2
`12(2)^2 + 12`
`48 + 12`
`60`
This coordinate will be `(2,60)`
Then, do the same thing to find the coordinates of the other point's you've chosen.

The points that I've chosen:
`(-4, 204) (-2,60) (2,60) (4,204)`
Notice how the `y` values for `-2` & `2` and `-4` & `4` are the same, even though the `x` values are different. This is because this graph is a parabola! (It's equation begins with `x^2` and will always be U-shaped. (Remember: V-shaped graphs are absolute value graphs and are completely different from parabolas!)

`f(x) = 12x^2 +12`
graph{12x^2+12 [-98.5, 84.7, -0.36, 91.24]}
(you can zoom in and out on the graph)

Hope this helps!