Question

Question #acd94

Answer 1

Vertex: `(-1, 0)`
Axis of symmetry: `x=-1`
"Max" value: `oo`
Range: (see answer for multiple ways to write it)

Explanation:

Axis of Symmetry
To find the axis of symmetry of a quadratic function in standard notation (`ax^2+bx+c`), you use `(-b)/(2a)`.

In the equation `x^2 + 2x + 1`,
`a=1`
`b=2`
`c=1`
So `(-b)/(2a)` = `(-2)/(2(1))` = `(-2)/2` = `-1`
Our axis of symmetry is `x=-1`.

Vertex
To find the vertex, take the axis of symmetry and plug it into the equation. (this will give us the `y`-value.)

So `(-1)^2 + 2(-1) + 1` = `1 + (-2) + 1` = `-1 + 1` = `0`

Our vertex, therefore, is on `(-1,0)`.

Max Value
First, we need to find out whether the graph is opening up or down.
Opening up means the parabola is like a curved V.
Opening down means the parabola is like a upside-down curved V.

Since our `a` value is positive, the parabola is opening up. Since the value of the equation keeps going up as `x` increases, there is no max value (the max value wouldn't be `oo` because it isn't technically a number)

Range
In this function, the minimum value for `y` is 0, and anything greater than or equal to 0 exists in the function (`y >= 0`)

Writing the range in set notation: `{y | y>= 0}`
You could also do: `{y in RR| y>= 0}`

Writing the range in interval notation: `[0,oo)`

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And for reference, here is what `x^2 + 2x + 1` would look like on a graph:
graph{x^2+2x+1 [-10, 10, -5, 5]}