The given expression is $f(x) = x^{2} + 8x + 7$, a quadratic function.

Domain:

Since this is a quadratic function (a polynomial), it is defined for all real values of $x$. There are no restrictions (like square roots of negative numbers or division by zero).

• Domain: $(-\infty, \infty)$ or all real numbers.

Range:

For the range, we need to determine the minimum or maximum value of the function, as quadratic functions are either concave up (minimum point) or concave down (maximum point). Since the coefficient of $x^2$ is positive, this parabola opens upwards, meaning it has a minimum value.

1. The vertex form of a quadratic is $f(x) = a(x-h)^2 + k$, where $(h, k)$ is the vertex.
2. To find the vertex, use the formula $x = -\frac{b}{2a}$, where $a = 1$ and $b = 8$.

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

Now, substitute $x = -4$ into the original function to find the $y$-coordinate of the vertex:

$f(-4) = (-4)^{2} + 8(-4) + 7 = 16 - 32 + 7 = -9$

Thus, the vertex is $(-4, -9)$, which means the minimum value of the function is $-9$.

• Range: $[-9, \infty)$

Summary:

• Domain: $(-\infty, \infty)$
• Range: $[-9, \infty)$

Do you have any questions or would you like more details?

Here are five related questions:

1. How do you convert a quadratic equation to vertex form?
2. What does the discriminant tell us about the roots of a quadratic equation?
3. How do you determine the axis of symmetry of a parabola?
4. Can quadratic functions have restricted domains?
5. How do you find the intercepts of a quadratic function?

Tip: In any quadratic equation $ax^2 + bx + c$, the parabola opens upwards if $a > 0$, and downwards if $a < 0$.