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Magazine Chapter 8: Problem 20
Graph each quadratic function, and state its domain and range. $$g(x)=x^{2}-4$$
Short Answer
Expert verified
Graph the parabola with vertex at \((0, -4)\). The domain is \((-\text{鈭瀩, \text{鈭瀩)\) and the range is \([-4, \text{鈭瀩)\).
Step by step solution
01
- Identify the Vertex
The function is in the form of a standard quadratic function, which is written as \(g(x)=x^2-4\). The vertex form of a quadratic function is \(y = a(x-h)^2 + k\), and here, the vertex \((h, k)\) is \((0, -4)\). This is because the equation can be simplified to \(g(x) = (x - 0)^2 - 4\), indicating \(h = 0\) and \(k = -4\).
02
- Determine the Direction of the Parabola
Since the coefficient of \(x^2\) is positive (\(+1\) in this case), the parabola opens upwards.
03
- Plot the Vertex
Plot the vertex \((0, -4)\) on the coordinate plane. This point is the minimum point of the parabola.
04
- Find Additional Points
To get a more accurate graph, choose additional x-values and find corresponding y-values by substituting into the function \(g(x) = x^2 - 4\). For example, let \(x = -2, -1, 1, 2\): \(g(-2) = (-2)^2 - 4 = 4 - 4 = 0\)\(g(-1) = (-1)^2 - 4 = 1 - 4 = -3\)\(g(1) = 1^2 - 4 = 1 - 4 = -3\)\(g(2) = 2^2 - 4 = 4 - 4 = 0\).
05
- Plot Additional Points
Plot the points \((-2, 0)\), \((-1, -3)\), \((1, -3)\), and \((2, 0)\) on the coordinate plane.
06
- Draw the Parabola
Draw a smooth curve through the vertex and all the additional points to form the parabola.
07
- State the Domain
For a quadratic function, the domain is all real numbers since there are no restrictions on the values of \(x\). Hence, the domain is \((-\text{鈭瀩, \text{鈭瀩)\).
08
- State the Range
Since the vertex is at the minimum point \((0, -4)\) and the parabola opens upwards, the range includes all real numbers greater than or equal to \(-4\). Hence, the range is \([-4, \text{鈭瀩)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vertex of a quadratic function
The vertex of a quadratic function is a crucial point on the graph because it represents either the highest or the lowest point of the parabola. For a function like \(g(x) = x^2 - 4\), you can find the vertex in a straightforward way. This function is in the standard form, \(y = ax^2 + bx + c\), where \(a = 1\), \(b = 0\), and \(c = -4\).
The vertex form of a quadratic function is \(y = a(x-h)^2 + k\). You can convert the standard form into the vertex form by completing the square, or simply recognize the structure when \(b = 0\). Here, it's clear that the vertex is at \((h, k) = (0, -4)\). This point represents the lowest point of the parabola since the parabola opens upwards (because \(a > 0\)). Remember, the vertex is always at the point \(x = \frac{-b}{2a}\) and \(y\) can be found by plugging this \(x\) back into the equation.
The vertex form of a quadratic function is \(y = a(x-h)^2 + k\). You can convert the standard form into the vertex form by completing the square, or simply recognize the structure when \(b = 0\). Here, it's clear that the vertex is at \((h, k) = (0, -4)\). This point represents the lowest point of the parabola since the parabola opens upwards (because \(a > 0\)). Remember, the vertex is always at the point \(x = \frac{-b}{2a}\) and \(y\) can be found by plugging this \(x\) back into the equation.
domain and range
The domain and range of a quadratic function tell you which values the function can take.
The domain is all the possible values for \(x\). For a quadratic function, there are no restrictions preventing \(x\) from being any real number. So, the domain is \((-\text{鈭瀩, \text{鈭瀩)\).
The range is all the possible values for \(y\). For \(g(x) = x^2 - 4\), you start by identifying the vertex. As we determined, the vertex is at \((0, -4)\). Since the parabola opens upwards, this means that the vertex represents the minimum \(y\) value. Therefore, \(y\) can be any value greater than or equal to \(-4\). So, the range is \([-4, \text{鈭瀩)\).
The domain is all the possible values for \(x\). For a quadratic function, there are no restrictions preventing \(x\) from being any real number. So, the domain is \((-\text{鈭瀩, \text{鈭瀩)\).
The range is all the possible values for \(y\). For \(g(x) = x^2 - 4\), you start by identifying the vertex. As we determined, the vertex is at \((0, -4)\). Since the parabola opens upwards, this means that the vertex represents the minimum \(y\) value. Therefore, \(y\) can be any value greater than or equal to \(-4\). So, the range is \([-4, \text{鈭瀩)\).
parabola direction
The direction of the parabola is determined by the coefficient \(a\) in the quadratic function's standard form \(y = ax^2 + bx + c\).
If \(a > 0\), the parabola opens upwards, meaning the arms of the parabola point up. This is because, as \(x^2\) becomes large, the entire expression \(ax^2\) increases without bound, pulling \(y\) upwards.
If \(a < 0\), the parabola opens downwards, meaning the arms of the parabola point down. In this case, as \(x^2\) becomes large, the expression \(ax^2\) decreases without bound, pushing \(y\) downwards.
For the function \(g(x) = x^2 - 4\), \(a = +1\), which is positive. So, this parabola opens upwards. This means the vertex at \((0, -4)\) is a minimum point, and as you move away from the vertex in either direction on the \(x\)-axis, the \(y\) values increase.
If \(a > 0\), the parabola opens upwards, meaning the arms of the parabola point up. This is because, as \(x^2\) becomes large, the entire expression \(ax^2\) increases without bound, pulling \(y\) upwards.
If \(a < 0\), the parabola opens downwards, meaning the arms of the parabola point down. In this case, as \(x^2\) becomes large, the expression \(ax^2\) decreases without bound, pushing \(y\) downwards.
For the function \(g(x) = x^2 - 4\), \(a = +1\), which is positive. So, this parabola opens upwards. This means the vertex at \((0, -4)\) is a minimum point, and as you move away from the vertex in either direction on the \(x\)-axis, the \(y\) values increase.
