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

Jun 12, 2016

The vertex is (2,3). X-intercepts are $2 \pm \sqrt{3.}$
Y-intercept = -1.

#### Explanation:

Let $y = f \left(x\right) = 3 - {\left(x - 2\right)}^{2}$
$\therefore y - 3 = - {\left(x - 2\right)}^{2}$

Clearly, the vertex is $\left(2 , 3\right)$

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

$y = 0$ $\Rightarrow - 3 = - {\left(x - 2\right)}^{2}$ $\Rightarrow$ $\left(x - 2\right) = \pm \sqrt{3}$
$\Rightarrow$ $x = 2 \pm \sqrt{3}$

So, X-intercepts are $2 \pm \sqrt{3}$

$x = 0$ $\Rightarrow$ $y - 3 = - {\left(0 - 2\right)}^{2} = - 4$, so $y = - 1$ is Y-intercept.