Question

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

Answer 1

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

`x_("intercepts")=1 and 3`
`y_("intercept")=-3`

increasing for `x<2`

Domain`->"input"-> {x: -oo<=x<=+oo}`

Explanation:

As the coefficient of `x^2` is -1 the graph is of form `nn`

Thus there is a determinable maximum.

Set `y=-(x-2)^2+1`

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`color(blue)("Determine the vertex")`

The given function is in the form `y=a(x+b/(2a))^2+k` which is the vertex form (completing the square). Thus with a slight modification the vertex may read off it directly.

`x_("vertex")=(-1)xxb/(2a)->(-1)xx(-2)=+2`

`y_("vertex")=k=1`

`color(green)(" Vertex"->(x,y)=(2,1))`
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`color(blue)("Determine x-intercepts")`

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

`y=0=-(x-2)^2+1`

Add `(x+2)^2` to both sides

`+(x-2)^2=1`

Square root both sides

`x-2 =+-1`

Add 2 to both sides

`color(green)(x=2+-1->x=1 and 3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine y-intercept")`

Set `x=0` giving

`y_("intercept")=-(0-2)^2+1`

`color(green)( y_("intercept")=-4+1=-3)`
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`color(blue)("Determine where the function is increasing")`

As the graph form is `nn` then `f(x)` is increasing to the left of the maximum at `(2,1)`. So increasing for `x<2`

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