The function $f(x) = 4x^2 + 64x + 262$. The function $g(x)$ is defined by $g(x) = f(x + 5)$. For what value of $x$ does $g(x)$ reach its minimum?

Neetesh Kumar | October 06, 2024

Math2A Course: Differential Equation

This is the solution to question related to Quadratic Equation
Contact me if you need help with Homework, Assignments, Tutoring Sessions, or Exams for STEM subjects.
You can see our Testimonials or Vouches from here of the previous works I have done.

Get Homework Help

---

Step-by-step solution:

Step 1: Find the minimum of $f(x)$

The function $f(x)$ is a quadratic function in the standard form $f(x) = ax^2 + bx + c$, where $a = 4$, $b = 64$, and $c = 262$.

The minimum (or maximum) of a quadratic function occurs at:

$x_{\text{min}} = \dfrac{-b}{2a}$

Substituting the values of $a$ and $b$:

$x_{\text{min}} = \frac{-64}{2(4)} = \frac{-64}{8} = -8$

Thus, the function $f(x)$ reaches its minimum at $x = -8$.

---

Step 2: Analyze the function $g(x)$

The function $g(x)$ is defined as $g(x) = f(x + 5)$. To find the minimum of $g(x)$, we need to shift the value of $x$ at which $f(x)$ reaches its minimum by $-5$ units.

Since $f(x)$ reaches its minimum at $x = -8$, $g(x)$ will reach its minimum when:

$x + 5 = -8$

Solving for $x$:

$x = -8 - 5 = -13$

The function $g(x)$ reaches its minimum at $x = -13$.

---

Conclusion:

Option - A $\checkmark$

Please comment below if you find any error in this solution.
If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions to similar questions.
Keep Smiling :-)