Math Problem Statement

f(x) = -(x + 1)2 + 4

Solution

The given function is:

f(x)=(x+1)2+4f(x) = -(x + 1)^2 + 4

This is a quadratic function, and the negative sign in front of (x+1)2(x + 1)^2 indicates that the parabola opens downward. Let's analyze the function step by step:

1. Vertex Form of a Quadratic Function

The function is already in vertex form, which is generally written as:

f(x)=a(xh)2+kf(x) = a(x - h)^2 + k

Where (h,k)(h, k) is the vertex of the parabola. For your function:

f(x)=(x+1)2+4f(x) = -(x + 1)^2 + 4

This is equivalent to:

f(x)=1(x(1))2+4f(x) = -1(x - (-1))^2 + 4

Thus, the vertex is at (1,4)(-1, 4).

2. Opening Direction

Since the coefficient of (x+1)2(x + 1)^2 is negative (i.e., 1-1), the parabola opens downward.

3. Y-intercept

To find the y-intercept, set x=0x = 0 and solve for f(x)f(x):

f(0)=(0+1)2+4=(1)2+4=1+4=3f(0) = -(0 + 1)^2 + 4 = -(1)^2 + 4 = -1 + 4 = 3

Thus, the y-intercept is f(0)=3f(0) = 3, or the point (0,3)(0, 3).

4. Axis of Symmetry

The axis of symmetry is the vertical line passing through the vertex. Since the vertex is at x=1x = -1, the axis of symmetry is:

x=1x = -1

5. X-intercepts (Roots)

To find the x-intercepts, set f(x)=0f(x) = 0:

0=(x+1)2+40 = -(x + 1)^2 + 4

Solving for xx:

(x+1)2=4(x + 1)^2 = 4

x+1=±2x + 1 = \pm 2

x=1±2x = -1 \pm 2

Thus, the x-intercepts are:

x=1+2=1andx=12=3x = -1 + 2 = 1 \quad \text{and} \quad x = -1 - 2 = -3

So, the x-intercepts are x=1x = 1 and x=3x = -3, or the points (1,0)(1, 0) and (3,0)(-3, 0).

Summary

  • Vertex: (1,4)(-1, 4)
  • Direction: Opens downward
  • Y-intercept: (0,3)(0, 3)
  • Axis of symmetry: x=1x = -1
  • X-intercepts: (1,0)(1, 0) and (3,0)(-3, 0)

Would you like more detailed explanations on any of these steps? Or perhaps visualizations of the function?

Related Questions:

  1. How would the graph change if the equation were f(x)=(x+1)2+4f(x) = (x + 1)^2 + 4?
  2. How to solve for the vertex of a quadratic function in general?
  3. What is the importance of the axis of symmetry in quadratic functions?
  4. How do you find the maximum or minimum value of a quadratic function?
  5. What are the different forms of quadratic functions, and how are they useful?

Tip: To sketch a quadratic function easily, always start by identifying the vertex, axis of symmetry, and x- and y-intercepts.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Vertex Form
Intercepts

Formulas

Vertex form of a quadratic function: f(x) = a(x - h)² + k
Intercepts calculation: Set f(x) = 0 or x = 0

Theorems

Properties of Parabolas
Axis of Symmetry Theorem

Suitable Grade Level

Grades 9-11