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Magazine Chapter 11: Problem 16
Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and sketch the graph. See Examples I through \(4 .\) (IMAGE CANNOT COPY) \(f(x)=-x^{2}+4 x-4\)
Short Answer
Expert verified
The vertex is (2, 0), the graph opens downward, the y-intercept is (0, -4), and it touches the x-axis at (2, 0).
Step by step solution
01
Identify the Coefficients
The quadratic function given is in the form \( f(x) = ax^2 + bx + c \). Here, the coefficients are \( a = -1 \), \( b = 4 \), and \( c = -4 \).
02
Determine the Vertex
The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. We find \( h \) using the formula \( h = \frac{-b}{2a} \). Substituting \( a = -1 \) and \( b = 4 \), we get:\[ h = \frac{-4}{2(-1)} = 2 \]Now substitute \( x = 2 \) into the function to find \( k \):\( f(2) = -(2)^2 + 4(2) - 4 = -4 + 8 - 4 = 0 \)Thus, the vertex is \( (2, 0) \).
03
Identify the Direction the Graph Opens
The coefficient \( a = -1 \) is negative, so the parabola opens downward.
04
Find the Y-Intercept
The y-intercept is found by setting \( x = 0 \) in the function. \[ f(0) = -0^2 + 4(0) - 4 = -4 \]Thus, the y-intercept is \( (0, -4) \).
05
Find the X-Intercepts
To find the x-intercepts, we set \( f(x) = 0 \) and solve the quadratic equation directly:\[ -x^2 + 4x - 4 = 0 \]Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = -1 \), \( b = 4 \), and \( c = -4 \):\[ x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(-4)}}{2(-1)} = \frac{-4 \pm \sqrt{16 - 16}}{-2} = \frac{-4 \pm 0}{-2} \]This results in a double root at \( x = 2 \), confirming symmetry at the vertex, so the parabola touches the x-axis at \( (2, 0) \).
06
Sketch the Graph
Combine the information gathered: The vertex is at \( (2, 0) \), the graph opens downward, the y-intercept is \( (0, -4) \), and there's a double root at the vertex \( (2, 0) \), indicating it only touches the x-axis there. Sketch the parabola opening downwards symmetrically around \( x = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A quadratic function is a type of polynomial function characterized by its highest degree being two, resulting in a graph in the shape of a parabola. The general form is expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
This structure forms the foundation of quadratic equations, which can represent various real-world phenomena, from the trajectory of thrown objects to economic models.
This structure forms the foundation of quadratic equations, which can represent various real-world phenomena, from the trajectory of thrown objects to economic models.
- The term \( ax^2 \) provides the distinctive parabolic shape.
- The term \( bx \) influences the location and orientation.
- The constant \( c \) moves the graph up or down along the y-axis.
Parabola Opens
The direction in which a parabola opens is determined by the coefficient \( a \) in the quadratic function \( f(x) = ax^2 + bx + c \). Whether the parabola opens upward or downward greatly affects the interpretation of the function.
- If \( a > 0 \), the parabola opens upward, resembling a smiling face. Here, the vertex represents the lowest point or the minimum of the function.
- If \( a < 0 \), the parabola opens downward, like a frown. In this configuration, the vertex serves as the highest point or the maximum.
X-Intercepts
The x-intercepts of a quadratic function are the points where the parabola crosses the x-axis. These intercepts occur where \( f(x) = 0 \). Solving the quadratic equation gives the x-intercepts, also known as roots or solutions.
For the quadratic \( f(x) = -x^2 + 4x - 4 \), the equation becomes \( -x^2 + 4x - 4 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gives us the intercepts.
For the quadratic \( f(x) = -x^2 + 4x - 4 \), the equation becomes \( -x^2 + 4x - 4 = 0 \). Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gives us the intercepts.
- When the discriminant \( \Delta = b^2 - 4ac \) is positive, there are two distinct x-intercepts.
- If \( \Delta \) equals zero, there's only one intercept, a double root, meaning the parabola touches the x-axis just once at its vertex.
- For \( \Delta < 0 \), there are no real x-intercepts as the parabola does not cross the x-axis.
Y-Intercept
The y-intercept is found where the function intersects the y-axis, occurring when \( x = 0 \). It provides a valuable reference point for understanding the graph's initial position and drawing it accurately.
To determine the y-intercept for a quadratic function \( f(x) = ax^2 + bx + c \), substitute \( x = 0 \) into the equation, leading to \( f(0) = c \). Thus, \( c \) is the y-intercept.
In our example, \( f(x) = -x^2 + 4x - 4 \), setting \( x = 0 \) gives us \( f(0) = -4 \).
This calculation shows the graph crosses the y-axis at the point \( (0, -4) \). When sketching or analyzing such graphs, knowing the y-intercept helps to position the parabola correctly.
To determine the y-intercept for a quadratic function \( f(x) = ax^2 + bx + c \), substitute \( x = 0 \) into the equation, leading to \( f(0) = c \). Thus, \( c \) is the y-intercept.
In our example, \( f(x) = -x^2 + 4x - 4 \), setting \( x = 0 \) gives us \( f(0) = -4 \).
This calculation shows the graph crosses the y-axis at the point \( (0, -4) \). When sketching or analyzing such graphs, knowing the y-intercept helps to position the parabola correctly.
