Question

How do you find the important points to graph `f(x)=2x^2`?

Answer 1

See explanation

Explanation:

Set `y=f(x)=2x^2`

Compare to the standardised form: `y=ax^2+bx+c`
This equation `color(white)("dddddfdd") ->color(white)("ddddd")y=2x^2+0x+0`

Key points:

`a>0 -> ("positive")` so the graph is of form `uu`

`c->` y-intercept `=0`

`x_("vertex") -> (-1/2)xx b/acolor(white)("ddd") =color(white)("ddd") (-1/2)xx0/2color(white)("d")=color(white)("d")0`
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As `x_("vertex")=0` the axis of symmetry is the y-axis

As `a>0` then the general shape is `uu` with the y-axis in the middle.

Couple this with `c=0` and it means that the vertex is at `(x,y)=(0.0)`