Certainly! Let's go through a step-by-step solution for the function [tex]\( f(x) = x^2 - 6x + 5 \)[/tex] to plot the required features on the coordinate plane.
### Step-by-Step Solution
1. Find the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts are the points where [tex]\( f(x) = 0 \)[/tex].
- Solving the quadratic equation [tex]\( x^2 - 6x + 5 = 0 \)[/tex], we get [tex]\( x = 5 \)[/tex] and [tex]\( x = 1 \)[/tex].
2. Find the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept is the point where [tex]\( x = 0 \)[/tex].
- Substituting [tex]\( x = 0 \)[/tex] into the function: [tex]\( f(0) = 0^2 - 6(0) + 5 = 5 \)[/tex].
- So, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 5) \)[/tex].
3. Find the vertex:
The vertex of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] is at [tex]\( x = -\frac{b}{2a} \)[/tex].
- Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -6 \)[/tex].
- So, the [tex]\( x \)[/tex]-coordinate is [tex]\( x = -\frac{-6}{2 \cdot 1} = 3 \)[/tex].
- Substituting [tex]\( x = 3 \)[/tex] into the function to find the [tex]\( y \)[/tex]-coordinate: [tex]\( f(3) = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4 \)[/tex].
- Hence, the vertex is [tex]\( (3, -4) \)[/tex].
4. Find the axis of symmetry:
The axis of symmetry for a parabola [tex]\( ax^2 + bx + c \)[/tex] is a vertical line through the vertex.
- So the axis of symmetry is the line [tex]\( x = 3 \)[/tex].
### Plotting the Points on the Coordinate Plane
- [tex]\( x \)[/tex]-intercepts: Points [tex]\( (5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
- [tex]\( y \)[/tex]-intercept: Point [tex]\( (0, 5) \)[/tex].
- Vertex: Point [tex]\( (3, -4) \)[/tex].
- Axis of symmetry: The vertical line [tex]\( x = 3 \)[/tex].
With this information, you can plot the points and draw the parabola as follows:
1. Mark the [tex]\( x \)[/tex]-intercepts at [tex]\( (5, 0) \)[/tex] and [tex]\( (1, 0) \)[/tex].
2. Mark the [tex]\( y \)[/tex]-intercept at [tex]\( (0, 5) \)[/tex].
3. Mark the vertex at [tex]\( (3, -4) \)[/tex].
4. Draw the axis of symmetry as a vertical dashed line passing through [tex]\( x = 3 \)[/tex].
5. Sketch the parabola opening upwards, passing through the marked intercepts and vertex, and symmetric around the line [tex]\( x = 3 \)[/tex].
By following these steps, you will have a well-constructed graph of the function [tex]\( f(x) = x^2 - 6x + 5 \)[/tex] with all the requested elements.
