The equation given in your problem is:

$y = x^2 + 8x + 18$

This is a quadratic equation representing a parabola. Let's go through the steps to find the vertex, axis of symmetry, minimum or maximum value, and the range.

1. Vertex:

To find the vertex of the parabola, we use the formula for the x-coordinate of the vertex in the quadratic equation $y = ax^2 + bx + c$, which is given by:

$x = -\frac{b}{2a}$

In your equation, $a = 1$ and $b = 8$.

$x = -\frac{8}{2(1)} = -4$

Now, substitute $x = -4$ back into the equation to find the y-coordinate:

$y = (-4)^2 + 8(-4) + 18 = 16 - 32 + 18 = 2$

So, the vertex is $(-4, 2)$.

2. Axis of Symmetry:

The axis of symmetry is the vertical line that passes through the x-coordinate of the vertex. So, the axis of symmetry is:

$x = -4$

3. Maximum or Minimum Value:

Since the parabola opens upwards (the coefficient of $x^2$ is positive), the vertex represents a minimum point. The minimum value of the function is the y-coordinate of the vertex, which is:

$y = 2$

4. Range:

Since the parabola opens upwards and has a minimum value at $y = 2$, the range of the function is:

$y \geq 2$

Summary:

• Vertex: $(-4, 2)$
• Axis of Symmetry: $x = -4$
• Minimum Value: $y = 2$
• Range: $y \geq 2$

Would you like more details or have any other questions? Here are some related questions to deepen your understanding:

1. How do you derive the formula for the x-coordinate of the vertex?
2. Can you rewrite the quadratic equation in vertex form?
3. How can you graph this parabola based on the information given?
4. What changes if the parabola opens downwards?
5. How do you find the intercepts of this quadratic function?

Tip: When working with parabolas, always check the sign of the $x^2$ term to determine whether the parabola opens upwards or downwards!