How do you find the vertex and the intercepts for $f(x)=3-(x-2)^2$ ?

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Answer 1 · 2016-06-12T19:37:42

The vertex is $(2,3).$ X-intercepts are $2+-sqrt3.$
Y-intercept = -1.

Let $y=f(x)=3-(x-2)^2$
$:. y-3= -(x-2)^2$

Clearly, the vertex is $(2,3)$

To find intercepts on X-axis & Y-axis, we have to take $y=0, x=0$ resp.

$y=0$ $rArr -3=-(x-2)^2$ $rArr$ $(x-2)=+-sqrt3$
$rArr$ $x=2+-sqrt3$

So, X-intercepts are $2+-sqrt3$

$x=0$ $rArr$ $y-3=-(0-2)^2 = -4$ , so $y=-1$ is Y-intercept.

How do you find the vertex and the intercepts for f(x)=3-(x-2)^2f(x)=3-(x-2)^2?