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Chapter 3: Problem 10

a. Determine whether the graph of the parabola b. Identify the vertex. opens upward or downward. c. Determine the \(x\) -intercept(s). d. Determine the \(y\) -intercept. e. Sketch the function. f. Determine the axis of symmetry. g. Determine the minimum or maximum value h. Determine the domain and range. (See Example 1) of the function. $$ g(x)=-(x+2)^{2}+4 $$

Short Answer

Expert verified
a. The parabola opens downward.b. The vertex is (-2, 4).c. The x-intercepts are -4 and 0.d. The y-intercept is 0.e. Graph with the vertex (-2, 4) and axis of symmetry x = -2.f. The axis of symmetry is x = -2.g. The maximum value is 4.h. Domain: (-鈭�, 鈭�), Range: (-鈭�, 4]

Step by step solution

01

Determine if the parabola opens upward or downward

The general form of a parabola is given by \( g(x) = ax^2 + bx + c \). If the coefficient of \( x^2 \) (which is \( a \)) is positive, the parabola opens upwards. If it is negative, the parabola opens downwards. Here, the function is \( g(x) = -(x+2)^2 + 4 \). Expand it to identify \( a \). \( g(x) = -(x^2 + 4x + 4) + 4 = -x^2 - 4x - 4 + 4 = -x^2 - 4x \). Thus, \( a = -1 \). Since \( a \) is negative, the parabola opens downward.
02

Identify the vertex

The function is in the vertex form \( g(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. For \( g(x) = -(x+2)^2 + 4 \), \( h = -2 \) and \( k = 4 \). So, the vertex is \( (-2, 4) \).
03

Determine the \( x \)-intercepts

To find the \( x \)-intercepts, set \( g(x) = 0 \) and solve for \( x \). \( 0 = -(x+2)^2 + 4 \). Let's solve it: \( (x+2)^2 = 4 \)\( x + 2 = \pm 2 \)\( x = 0 \pm 2 - 2 \)\( x = 0 \pm 2 - 2 \)or\( x = -4 \, or \, x = 0 \)
04

Determine the \( y \)-intercept

To find the \( y \)-intercept, evaluate \( g(x) \) at \( x = 0 \). \( g(0) = -(0+2)^2 + 4 = -4 + 4 = 0 \). So, the \( y \)-intercept is \( (0, 0) \).
05

Determine the axis of symmetry

The axis of symmetry of the parabola is \( x = h \). Here, \( h = -2 \). Therefore, the axis of symmetry is \( x = -2 \).
06

Determine the minimum or maximum value

Since the parabola opens downward, the vertex represents the maximum value. The maximum value is \( k = 4 \).
07

Determine the domain and range

The domain of any quadratic function is all real numbers, i.e., \( (-\infty, \infty) \). The range is from the maximum value downwards. Hence, the range is \( (-\infty, 4] \).
08

Sketch the function

Plot the vertex and axis of symmetry. Draw the parabola opening downward through the intercepts. The vertex is (-2, 4) and it passes through the points (0, 0) and (-4, 0). The axis of symmetry is x = -2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

vertex identification
The vertex of a parabola is a key point that tells us the highest or lowest point of the graph. For any quadratic function written in the form \( g(x) = a(x-h)^2 + k \), the vertex is \( (h, k) \). Here, \( g(x) = -(x+2)^2 + 4 \), which fits the vertex form with \( h = -2 \) and \( k = 4 \). Thus, the vertex of the parabola is at \( (-2, 4) \). This means the highest point on this graph is at \( (-2, 4) \) because the parabola opens downward.
parabola direction
The direction a parabola opens depends on the coefficient of the \( x^2 \) term in the quadratic equation \( g(x) = ax^2 + bx + c \). If \( a \) is positive, the parabola opens upward. If \( a \) is negative, it opens downward. In this example, \( g(x) = -(x+2)^2 + 4 \), the coefficient of the \( x^2 \) term is \( -1 \). Since \( -1 \) is negative, the parabola opens downward. This affects where the maximum value of the function is found.
x-intercepts
The \( x \) - intercepts of a parabola are the points where the graph crosses the \( x \) -axis. To find these points, set \( g(x) \) to 0 and solve for \( x \). For the equation \( g(x) = -(x+2)^2 + 4 \):
\( 0 = -(x+2)^2 + 4 \)
Solving for \( x \):
\( (x+2)^2 = 4 \)
\( x+2 = \pm 2 \)
So, \( x = -4 \) or \( x = 0 \).
These are the \( x \)-intercepts: \( (-4, 0) \) and \( (0, 0) \).
y-intercepts
The \( y \) - intercept is the point where the graph crosses the \( y \) -axis. This happens where \( x = 0 \). To find the \( y \) - intercept for the equation \( g(x) = -(x+2)^2 + 4 \), substitute \( x = 0 \):
\( g(0) = -(0+2)^2 + 4 \)
\( g(0) = -4 + 4 = 0 \)
This means the \( y \) - intercept is at \( (0, 0) \).
axis of symmetry
The axis of symmetry of a parabola is a vertical line that goes through the vertex and divides the parabola into two mirror images. For the vertex form \( g(x) = a(x-h)^2 + k \), the axis of symmetry is \( x = h \). Here, \( h = -2 \), so the axis of symmetry is \( x = -2 \). This line helps visually balance the parabola.
maximum value
Since the parabola \( g(x) = -(x+2)^2 + 4 \) opens downward, the vertex represents the maximum value, which is the highest point on the graph. We know from the vertex \( (h, k) \) that \( k = 4 \). Therefore, the maximum value of the parabola is \( 4 \). This means that the output value of the function never goes above \( 4 \).
domain and range
The domain of a function is the set of all possible input values (or \( x \) values). For any quadratic function, the domain is all real numbers, \( (-\infty, \infty) \). The range is all possible output values (or \( y \) values).
Since the maximum value of \( g(x) \) is \( 4 \) and the parabola opens downward, the range is all real values less than or equal to \( 4 \). So, the range is \( (-\infty, 4] \).
function sketching
To sketch the function \( g(x) = -(x+2)^2 + 4 \), follow these steps:
  • Plot the vertex \( (-2, 4) \).
  • Draw the axis of symmetry \( x = -2 \).
  • Identify and plot the \( x \)-intercepts \( (-4, 0) \) and \( (0, 0) \).
  • Mark the \( y \)-intercept at \( (0, 0) \).
  • Sketch the parabola opening downward through these points.
This helps visualize how the function behaves across the coordinate plane.

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