How do you graph $f(x)=3-x^2$ and identify the x intercepts, vertex?

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Answer 1 · 2017-07-26T17:26:39

Use the standard graph of $y=x^2$ but upside down $(y=-x^2)$ and then shift it up 3 units.

To find the $x$ intercepts, make $f(x)=0$ so $x^2=3$ . Taking the square root of both sides gives $x=+-sqrt3$ .

The vertex of the graph will be at $(0,3)$ because the graph hasn't been translated left or right so is still symmetrical about the y-axis.

How do you graph f(x)=3-x^2f(x)=3-x^2 and identify the x intercepts, vertex?