Graph [tex]$s(x)=x^2-2x+4$[/tex] on the Cartesian plane to the right.

To graph the function, you'll first have to plot the vertex (high/low point), then any other point.

1. [tex]s(x)[/tex] is increasing on the interval(s):
[tex]\[
(1, \infty)
\][/tex]

2. [tex]s(x)[/tex] is decreasing on the interval(s):
[tex]\[
(-\infty, 1)
\][/tex]

3. [tex]s(x)[/tex] has a maximum of DNE at [tex]x= \text{DNE} \, \square[/tex]

4. [tex]s(x)[/tex] has a minimum of 3 at [tex]x=1 \, \square[/tex]

5. The domain of [tex]s(x)[/tex] is:
[tex]\[
(-\infty, \infty) \, \square
\][/tex]

6. The range of [tex]s(x)[/tex] is:
[tex]\[
[3, \infty) \, \square
\][/tex]



Answer :

Sure, let's graph and analyze the function [tex]\( s(x) = x^2 - 2x + 4 \)[/tex].

1. Identify the Vertex:

The vertex form of a parabola [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the vertex [tex]\( x = -\frac{b}{2a} \)[/tex].

For [tex]\( s(x) = x^2 - 2x + 4 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]

So the x-coordinate of the vertex is:
[tex]\[ x = -\frac{-2}{2 \cdot 1} = 1 \][/tex]

Substitute [tex]\( x = 1 \)[/tex] back into the function to find the y-coordinate of the vertex:
[tex]\[ s(1) = (1)^2 - 2(1) + 4 = 1 - 2 + 4 = 3 \][/tex]

Thus, the vertex is [tex]\( (1, 3) \)[/tex].

2. Plot the Vertex:

On the Cartesian plane, plot the point [tex]\( (1, 3) \)[/tex].

3. Find Additional Points:

For better understanding, let's choose another x-value, for instance, [tex]\( x = 0 \)[/tex].

Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ s(0) = (0)^2 - 2(0) + 4 = 4 \][/tex]

So another point on the graph is [tex]\( (0, 4) \)[/tex].

Additionally, choose [tex]\( x = 2 \)[/tex]:
[tex]\[ s(2) = (2)^2 - 2(2) + 4 = 4 - 4 + 4 = 4 \][/tex]

So another point on the graph is [tex]\( (2, 4) \)[/tex].

4. Sketch the Parabola:

Using the vertex [tex]\( (1, 3) \)[/tex] and points [tex]\( (0, 4) \)[/tex] and [tex]\( (2, 4) \)[/tex], sketch the parabola opening upwards.

5. Analyze the Function:

- Intervals of Increasing and Decreasing:
- The function [tex]\( s(x) \)[/tex] is a parabola that opens upwards.
- It is decreasing on [tex]\( (-\infty, 1) \)[/tex] (as [tex]\( x \)[/tex] moves towards 1 from the left).
- It is increasing on [tex]\( (1, \infty) \)[/tex] (as [tex]\( x \)[/tex] moves away from 1 to the right).

- Maximum or Minimum:
- Since the parabola opens upwards, [tex]\( s(x) \)[/tex] has no maximum.
- It has a minimum value at the vertex.
- The minimum value is 3 at [tex]\( x = 1 \)[/tex].

- Domain and Range:
- The domain of [tex]\( s(x) \)[/tex], since it is a quadratic function, is [tex]\( (-\infty, \infty) \)[/tex].
- The range, given that the parabola opens upwards and the minimum value is 3, is [tex]\( [3, \infty) \)[/tex].

To summarize:

- Vertex: [tex]\( (1, 3) \)[/tex]
- Increasing on: [tex]\( (1, \infty) \)[/tex]
- Decreasing on: [tex]\( (-\infty, 1) \)[/tex]
- Maximum: Does not exist (DNE)
- Minimum: 3 at [tex]\( x = 1 \)[/tex]
- Domain: [tex]\( (-\infty, \infty) \)[/tex]
- Range: [tex]\( [3, \infty) \)[/tex]