Math Problem Statement

solve

Solution

The problem given is:

f(x)=2x210x+5f(x) = 2x^2 - 10x + 5

The function g(x)g(x) is defined as g(x)=f(x+3)g(x) = f(x+3). We need to find the value of xx for which g(x)g(x) reaches its minimum.

Step 1: Express g(x)g(x) in terms of xx

To do this, substitute x+3x + 3 into the function f(x)f(x):

g(x)=f(x+3)=2(x+3)210(x+3)+5g(x) = f(x + 3) = 2(x + 3)^2 - 10(x + 3) + 5

Step 2: Expand and simplify the expression for g(x)g(x)

Expand (x+3)2(x + 3)^2:

(x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9

Now substitute this back into the equation for g(x)g(x):

g(x)=2(x2+6x+9)10(x+3)+5g(x) = 2(x^2 + 6x + 9) - 10(x + 3) + 5

Distribute the terms:

g(x)=2x2+12x+1810x30+5g(x) = 2x^2 + 12x + 18 - 10x - 30 + 5

Combine like terms:

g(x)=2x2+2x7g(x) = 2x^2 + 2x - 7

Step 3: Find the minimum value of g(x)g(x)

The minimum value of a quadratic function ax2+bx+cax^2 + bx + c occurs at x=b2ax = -\frac{b}{2a}. Here, a=2a = 2 and b=2b = 2:

xmin=22×2=24=12x_{\text{min}} = -\frac{2}{2 \times 2} = -\frac{2}{4} = -\frac{1}{2}

So, the minimum value of g(x)g(x) occurs at x=12x = -\frac{1}{2}.

Step 4: Verify with the answer choices

The answer choices provided were integers, but our calculated minimum point is x=12x = -\frac{1}{2}, which does not correspond directly to any of the choices.

This discrepancy suggests a mistake might have been made in expanding the function or interpreting the problem. Let's check it again to ensure correctness:

  • We correctly applied the formula for the vertex of the parabola.
  • However, the exact minimum value might be needed to compare against the options, or it's possible the problem's structure demands an adjustment in interpretation.

Final Answer

Given the above, it seems the options might represent the y-value (the minimum point) rather than the x-value, or there might be a misinterpretation of the question. I would recommend reconsidering the problem statement to match the answer choices provided.

Let me know if you need further details or have any questions!


  1. How do you determine the vertex of a quadratic function?
  2. Can the minimum value be found using derivatives?
  3. How does completing the square help in finding the vertex?
  4. What are the implications of translating a function horizontally?
  5. How do you confirm if a function opens upwards or downwards?

Tip: When dealing with quadratic functions, always check if you're solving for the vertex (x-coordinate) or the actual minimum/maximum value (y-coordinate).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Vertex of a Parabola
Minimum of a Quadratic Function

Formulas

Vertex formula for a quadratic function

Theorems

Vertex formula theorem

Suitable Grade Level

Grades 10-12