Let's work through the problem presented on the screen step by step.

The given quadratic function is:

$y = x^2 - 10x + 28$

a. Construct a table of values with 5 ordered pairs

We can calculate the values of $y$ for various values of $x$. Let's select $x = 0, 2, 4, 6, 8$ for simplicity.

Thus, the ordered pairs are:
$(0, 28), (2, 4), (4, 0), (6, 4), (8, 16)$

b. Sketch the graph of the function

• Plot the points from the table: $(0, 28), (2, 4), (4, 0), (6, 4), (8, 16)$.
• Connect these points in a smooth parabolic shape.

c to i. Identify the following characteristics:

c. Direction of the Opening of the Graph

Since the coefficient of $x^2$ is positive ($1$), the parabola opens upward.

d. Vertex

The vertex of a parabola given by $y = ax^2 + bx + c$ can be calculated using the formula:

$x_{\text{vertex}} = \frac{-b}{2a}$

For $y = x^2 - 10x + 28$, $a = 1$ and $b = -10$:

$x_{\text{vertex}} = \frac{-(-10)}{2(1)} = 5$

Substitute $x = 5$ back into the equation to find the $y$-coordinate of the vertex:

$y_{\text{vertex}} = 5^2 - 10(5) + 28 = 25 - 50 + 28 = 3$

So, the vertex is at $(5, 3)$.

e. Domain

For any quadratic function, the domain is all real numbers:

$\text{Domain: } (-\infty, \infty)$

f. Range

Since the parabola opens upward and its minimum point is the vertex at $(5, 3)$, the range is:

$\text{Range: } [3, \infty)$

g. $x$-intercepts

To find the $x$-intercepts, set $y = 0$:

$0 = x^2 - 10x + 28$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(28)}}{2(1)}$ $x = \frac{10 \pm \sqrt{100 - 112}}{2}$ $x = \frac{10 \pm \sqrt{-12}}{2}$

Since the discriminant is negative, the function has no real $x$-intercepts.

h. $y$-intercept

To find the $y$-intercept, set $x = 0$:

$y = 0^2 - 10(0) + 28 = 28$

So, the $y$-intercept is at $(0, 28)$.

i. Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. Since the vertex is at $x = 5$, the equation of the axis of symmetry is:

$x = 5$

Would you like more details on any of these steps?

Related Questions:

1. How would you solve a quadratic equation with real $x$-intercepts?
2. How can you determine the vertex of a quadratic function from its standard form?
3. What are the differences between the domain and range of a quadratic function?
4. How do you solve for the $x$-intercepts using the quadratic formula?
5. What is the significance of the axis of symmetry in a parabola?

Tip:

To quickly find the direction of a parabola, look at the sign of the coefficient of $x^2$: positive means it opens upward, negative means it opens downward.