Problem 16
Question
Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=-x^{2}+4 x-4 $$
Step-by-Step Solution
Verified Answer
The vertex is (2, 0), the graph opens downward, y-intercept is (0, -4), and x-intercept is (2, 0).
1Step 1: Determine the Form of the Quadratic Function
The given function is \( f(x) = -x^2 + 4x - 4 \). This is in the standard quadratic form \( ax^2 + bx + c \) where \( a = -1 \), \( b = 4 \), and \( c = -4 \).
2Step 2: Find the Vertex
The vertex form of a quadratic function is \( f(x) = a(x-h)^2 + k \), and the vertex \((h, k)\) can be found using the formula \( h = -\frac{b}{2a} \). Substitute \( b = 4 \) and \( a = -1 \): \[ h = -\frac{4}{2(-1)} = 2 \]Now substitute \( x = 2 \) back into the function to find \( k \):\[ f(2) = -2^2 + 4(2) - 4 = -4 + 8 - 4 = 0 \]The vertex is \((2, 0)\).
3Step 3: Determine Opening Direction
Since \( a = -1 \) in the function \( f(x) = -x^2 + 4x - 4 \), the graph opens downward because \( a < 0 \).
4Step 4: Find the Intercepts
To find the y-intercept, set \( x = 0 \): \[ f(0) = -(0)^2 + 4(0) - 4 = -4 \]The y-intercept is \( (0, -4) \).For the x-intercepts, set \( f(x) = 0 \):\[ 0 = -x^2 + 4x - 4 \]This simplifies to \( x^2 - 4x + 4 = 0 \), and factoring gives \((x-2)^2 = 0 \) which implies \( x = 2 \).Thus, the x-intercept is \((2, 0)\).
5Step 5: Graph the Function
Plot the vertex \((2, 0)\), the y-intercept \((0, -4)\), and note the graph opens downward. Since \((2, 0)\) is also the x-intercept, plot the parabola opening downwards from the vertex, with the axis of symmetry at \( x = 2 \).
Key Concepts
Vertex of a ParabolaIntercepts of a QuadraticQuadratic Graph Analysis
Vertex of a Parabola
The vertex of a parabola is a key point that provides a lot of information about the quadratic function. In this specific quadratic function, \( f(x) = -x^2 + 4x - 4 \), the vertex can be found by using the formula \( h = -\frac{b}{2a} \). By plugging in the values \( b = 4 \) and \( a = -1 \), we calculate \( h = 2 \). This means the x-value of the vertex is 2.
Next, substitute this x-value back into the function to find the y-coordinate of the vertex. When \( x = 2 \), the calculation \( f(2) = -2^2 + 4(2) - 4 = 0 \) gives us the y-value. Thus, the vertex of the parabola is \((2, 0)\).
This point is crucial for graphing the quadratic as it signifies the maximum or minimum of the function. In this case, because the coefficient \( a = -1 \) is negative, the parabola opens downward, making the vertex a high point or maximum in the graph.
Next, substitute this x-value back into the function to find the y-coordinate of the vertex. When \( x = 2 \), the calculation \( f(2) = -2^2 + 4(2) - 4 = 0 \) gives us the y-value. Thus, the vertex of the parabola is \((2, 0)\).
This point is crucial for graphing the quadratic as it signifies the maximum or minimum of the function. In this case, because the coefficient \( a = -1 \) is negative, the parabola opens downward, making the vertex a high point or maximum in the graph.
Intercepts of a Quadratic
Intercepts are the points where a graph crosses the axes. For the quadratic function \( f(x) = -x^2 + 4x - 4 \), we start by finding the y-intercept by setting \( x = 0 \). This results in \( f(0) = -4 \), so the y-intercept is at \((0, -4)\). This shows where the parabola meets the y-axis.
Next, we look for the x-intercepts where \( f(x) = 0 \). Solving the equation \( -x^2 + 4x - 4 = 0 \), we rearrange it to \( x^2 - 4x + 4 = 0 \). Factoring gives \((x-2)^2 = 0 \), indicating a double root at \( x = 2 \). Therefore, the x-intercept is at \((2, 0)\).
These intercepts provide additional key points that help define the shape and position of the parabola on the graph.
Next, we look for the x-intercepts where \( f(x) = 0 \). Solving the equation \( -x^2 + 4x - 4 = 0 \), we rearrange it to \( x^2 - 4x + 4 = 0 \). Factoring gives \((x-2)^2 = 0 \), indicating a double root at \( x = 2 \). Therefore, the x-intercept is at \((2, 0)\).
These intercepts provide additional key points that help define the shape and position of the parabola on the graph.
Quadratic Graph Analysis
Analyzing the graph of a quadratic function involves understanding its shape and the information provided by its vertex and intercepts. For \( f(x) = -x^2 + 4x - 4 \), identifying these key points helps us graph the parabola accurately.
The vertex \((2, 0)\) tells us the graph's highest point, acting as a peak since the parabola opens downward due to a negative leading coefficient \( a = -1 \). This vertex also serves as one of the x-intercepts.
The graph is symmetrical about the vertical line \( x = 2 \), which is the axis of symmetry. Understanding this symmetry aids in predicting the graph's shape and direction.
The vertex \((2, 0)\) tells us the graph's highest point, acting as a peak since the parabola opens downward due to a negative leading coefficient \( a = -1 \). This vertex also serves as one of the x-intercepts.
The graph is symmetrical about the vertical line \( x = 2 \), which is the axis of symmetry. Understanding this symmetry aids in predicting the graph's shape and direction.
- The parabola crosses the y-axis at \((0, -4)\), providing a starting point for sketching.
- Knowing the x-intercept at \((2, 0)\) allows us to plot the graph tracing back to the vertex.
- The defined vertex and intercepts enable us to draw a smooth curve representing the quadratic's path.
Other exercises in this chapter
Problem 16
Solve. Write the solution set in interval notation. $$ \frac{x-5}{x-6}>0 $$
View solution Problem 16
Solve. See Example 3. $$ x^{4}+2 x^{2}-3=0 $$
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Use the square root property to solve each equation. $$ x^{2}+4=0 $$
View solution Problem 16
Graph each quadratic function. Label the vertex and sketch and label the axis of symmetry. \(G(x)=(x+3)^{2}+3\)
View solution