Math Problem Statement

Write the expression f(x) = 2x^2 + 12x + 14 in the form f(x) = a(x + h)^2 + k and hence state the turning point of f(x). Calculate the value(s) of x that are valid for the equation |(x - 2) / (x + 3)| = 4.

Solution

Let's break down the two parts of this question.

Part (c): Rewrite f(x)=2x2+12x+14f(x) = 2x^2 + 12x + 14 in the form f(x)=a(x+h)2+kf(x) = a(x + h)^2 + k

This is asking us to complete the square.

  1. Start with: f(x)=2x2+12x+14f(x) = 2x^2 + 12x + 14
  2. Factor out 2 from the first two terms: f(x)=2(x2+6x)+14f(x) = 2(x^2 + 6x) + 14
  3. Complete the square inside the parentheses. Take half of 6, square it, and add/subtract it: x2+6x=(x+3)29x^2 + 6x = (x + 3)^2 - 9 So: f(x)=2((x+3)29)+14f(x) = 2\left((x + 3)^2 - 9\right) + 14
  4. Expand and simplify: f(x)=2(x+3)218+14=2(x+3)24f(x) = 2(x + 3)^2 - 18 + 14 = 2(x + 3)^2 - 4 Thus, the expression in completed square form is: f(x)=2(x+3)24f(x) = 2(x + 3)^2 - 4

The parameters are:

  • a=2a = 2, h=3h = -3, and k=4k = -4.

Turning point:

Since the function is written as f(x)=a(x+h)2+kf(x) = a(x + h)^2 + k, the turning point is given by: Turning point=(h,k)=((3),4)=(3,4)\text{Turning point} = (-h, k) = (-(-3), -4) = (3, -4)


Part (d): Solve x2x+3=4\left| \frac{x - 2}{x + 3} \right| = 4

This equation involves an absolute value. To solve it, we break it into two cases.

Case 1: x2x+3=4\frac{x - 2}{x + 3} = 4

x2=4(x+3)x - 2 = 4(x + 3) x2=4x+12x - 2 = 4x + 12 212=4xx14=3x-2 - 12 = 4x - x \Rightarrow -14 = 3x x=143x = -\frac{14}{3}

Case 2: x2x+3=4\frac{x - 2}{x + 3} = -4

x2=4(x+3)x - 2 = -4(x + 3) x2=4x12x - 2 = -4x - 12 x+4x=12+25x=10x + 4x = -12 + 2 \Rightarrow 5x = -10 x=2x = -2

Thus, the solutions to the equation are: x=143andx=2x = -\frac{14}{3} \quad \text{and} \quad x = -2


Do you have any questions or would you like further clarification on these steps?


Here are 5 related questions to explore:

  1. How do you find the vertex of a quadratic function without completing the square?
  2. What does the value of aa indicate about the shape of a parabola?
  3. How would the solutions change if the absolute value equation in part (d) had a different constant on the right-hand side?
  4. Can a quadratic function have more than one turning point? Why or why not?
  5. How do you solve inequalities involving absolute values?

Tip: Always double-check your solutions when dealing with absolute value equations, as extraneous solutions can arise.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Completing the Square
Absolute Value Equations

Formulas

Standard form of a quadratic function f(x) = a(x + h)^2 + k
Absolute value equation |(x - 2) / (x + 3)| = 4

Theorems

Properties of quadratic functions
Absolute value properties

Suitable Grade Level

Grades 10-12