# How do you find the domain, find where f is increasing, vertex, intercepts, maximum of f(x)=-(x-2)^2+1?

Jul 17, 2017

Vertex->(x,y)=(2,1)) larr" A maximum"

${x}_{\text{intercepts}} = 1 \mathmr{and} 3$
${y}_{\text{intercept}} = - 3$

increasing for $x < 2$

Domain$\to \text{input} \to \left\{x : - \infty \le x \le + \infty\right\}$

#### Explanation:

As the coefficient of ${x}^{2}$ is -1 the graph is of form $\cap$

Thus there is a determinable maximum.

Set $y = - {\left(x - 2\right)}^{2} + 1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the vertex}}$

The given function is in the form $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k$ which is the vertex form (completing the square). Thus with a slight modification the vertex may read off it directly.

${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{2 a} \to \left(- 1\right) \times \left(- 2\right) = + 2$

${y}_{\text{vertex}} = k = 1$

$\textcolor{g r e e n}{\text{ Vertex} \to \left(x , y\right) = \left(2 , 1\right)}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine x-intercepts}}$

These occur at $y = 0$ so we have:

$y = 0 = - {\left(x - 2\right)}^{2} + 1$

Add ${\left(x + 2\right)}^{2}$ to both sides

$+ {\left(x - 2\right)}^{2} = 1$

Square root both sides

$x - 2 = \pm 1$

$\textcolor{g r e e n}{x = 2 \pm 1 \to x = 1 \mathmr{and} 3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine y-intercept}}$

Set $x = 0$ giving

${y}_{\text{intercept}} = - {\left(0 - 2\right)}^{2} + 1$

$\textcolor{g r e e n}{{y}_{\text{intercept}} = - 4 + 1 = - 3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine where the function is increasing}}$

As the graph form is $\cap$ then $f \left(x\right)$ is increasing to the left of the maximum at $\left(2 , 1\right)$. So increasing for $x < 2$

It is neither increasing or decreasing at $x = 2$