Question

cc.2.2.hs.d.7
what is the range of the function $f(x) = x^2 + 8x - 3$?
$\big\\{f(x)\big|f(x) \geq -19\big\\}$
$\big\\{f(x)\big|f(x) \geq -11\big\\}$
$\big\\{f(x)\big|f(x) \geq -8\big\\}$
$\big\\{f(x)\big|f(x) \geq -3\big\\}$

Explanation:

Step1: Complete the square for the quadratic function

To find the range of \( f(x) = x^2 + 8x - 3 \), we complete the square. The coefficient of \( x \) is 8, half of it is 4, and squaring it gives 16. So we rewrite the function as:
\[

$$\begin{align*} f(x)&=x^2 + 8x + 16 - 16 - 3\\ &=(x + 4)^2 - 19 \end{align*}$$

\]

Step2: Analyze the range based on the squared term

Since the square of any real number \( (x + 4)^2 \) is always non - negative, i.e., \( (x + 4)^2\geq0 \) for all real numbers \( x \).
If we add - 19 to both sides of the inequality \( (x + 4)^2\geq0 \), we get \( (x + 4)^2-19\geq - 19 \).
Since \( f(x)=(x + 4)^2-19 \), this means \( f(x)\geq - 19 \).

Answer:

\(\{f(x)|f(x)\geq - 19\}\) (the first option)