To graph the function $y = x^2 - bx + 5$, we can explore the effect of the parameter $b$ on the graph's shape and position.

1. General Shape: The function $y = x^2 - bx + 5$ is a quadratic equation, so its graph will be a parabola opening upwards, as the coefficient of $x^2$ is positive.

2. Vertex: The vertex of a parabola given by $y = ax^2 + bx + c$ is located at $x = -\frac{b}{2a}$. Here, $a = 1$ and $c = 5$, so: $x = -\frac{-b}{2 \cdot 1} = \frac{b}{2}.$ Therefore, the x-coordinate of the vertex depends on the value of $b$, affecting the horizontal location of the parabola’s vertex.

3. Intercepts:

   • Y-Intercept: At $x = 0$, $y = 5$, so the y-intercept is always $(0, 5)$.
   • X-Intercepts: To find x-intercepts, we solve $x^2 - bx + 5 = 0$. The solutions (roots) will depend on $b$ and are given by the quadratic formula: $x = \frac{b \pm \sqrt{b^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{b \pm \sqrt{b^2 - 20}}{2}.$ The discriminant $b^2 - 20$ will determine the number of real roots:
     • If $b^2 > 20$, there are two distinct real roots.
     • If $b^2 = 20$, there is one real root (the vertex touches the x-axis).
     • If $b^2 < 20$, there are no real x-intercepts.

4. Effect of Changing $b$: Changing $b$ shifts the parabola horizontally and affects the vertex and the roots (if they exist). Higher values of $|b|$ shift the vertex further left or right.

Would you like me to plot the graph for a specific value of $b$?

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Here are some additional questions for practice:

1. What happens to the graph if $b = 0$?
2. How does changing the value of $c$ in a quadratic function $y = x^2 - bx + c$ affect its graph?
3. For which values of $b$ will the function have no real x-intercepts?
4. How would the graph change if the equation were $y = -x^2 - bx + 5$?
5. Can you find the coordinates of the vertex for $b = 4$?

Tip: Remember that the discriminant in a quadratic equation, $b^2 - 4ac$, helps determine the nature and number of roots.