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Magazine Chapter 8: Problem 80
Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find the \(y\) -intercept, approximate the \(x\) -intercepts to one decimal place, and sketch the graph. $$ f(x)=x^{2}-6 x+4 $$
Short Answer
Expert verified
Vertex: (3, -5); Upward; y-intercept: (0, 4); x-intercepts: (1.6, 0), (4.4, 0).
Step by step solution
01
Identifying the Form
The given quadratic is in standard form: \( f(x) = ax^2 + bx + c \). Here, \( a = 1 \), \( b = -6 \), and \( c = 4 \).
02
Finding the Vertex
The vertex of a parabola in the form \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values: \( x = -\frac{-6}{2\times1} = 3 \). Now find \( y \) by plugging \( x = 3 \) back into the function: \( f(3) = 3^2 - 6\times3 + 4 = -5 \). Thus, the vertex is \( (3, -5) \).
03
Determining the Direction
The direction in which the parabola opens is determined by the sign of \( a \). Since \( a = 1 > 0 \), the parabola opens upward.
04
Finding the y-intercept
The \( y \)-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function: \( f(0) = 0^2 - 6\times0 + 4 = 4 \). Thus, the \( y \)-intercept is \( (0, 4) \).
05
Finding the x-intercepts
Solve for \( x \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 4}}{2 \times 1} \]. This simplifies to \( x = \frac{6 \pm \sqrt{20}}{2} \). Approximating \( x \), we get \( x \approx 1.6 \) and \( x \approx 4.4 \). The \( x \)-intercepts are approximately \( (1.6, 0) \) and \( (4.4, 0) \).
06
Graph the Function
Plot the vertex \((3, -5)\), the y-intercept \((0, 4)\), and the x-intercepts \((1.6, 0)\) and \((4.4, 0)\) on a coordinate grid. Draw a parabola through these points opening upward.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex of a Parabola
The vertex is a key part of understanding quadratic functions, often described as the "turning point" of the graph. For a parabola given by the equation \( ax^2 + bx + c \), the vertex can be calculated using the formula \( x = -\frac{b}{2a} \). This formula helps find the x-coordinate of the vertex.
Plug this value back into the function to determine the y-coordinate. For the function \( f(x) = x^2 - 6x + 4 \), calculate the vertex:
Plug this value back into the function to determine the y-coordinate. For the function \( f(x) = x^2 - 6x + 4 \), calculate the vertex:
- First, identify \( a = 1 \), \( b = -6 \), and \( c = 4 \).
- Use the formula to find \( x = -\frac{-6}{2 \times 1} = 3 \).
- Substitute \( x = 3 \) back into the function to find the y-coordinate: \( f(3) = 3^2 - 6 \times 3 + 4 = -5 \).
y-intercept
The y-intercept of a quadratic function is the point where the graph crosses the y-axis. This occurs when the x-coordinate is zero, simplifying the process of finding it. For any quadratic equation in standard form \( f(x) = ax^2 + bx + c \), the y-intercept is given by \( c \).
In our example, \( f(x) = x^2 - 6x + 4 \), the y-intercept is found by setting \( x = 0 \) and solving the equation:
This point is essential for graphing as it provides a definitive starting point where the parabola crosses the y-axis.
In our example, \( f(x) = x^2 - 6x + 4 \), the y-intercept is found by setting \( x = 0 \) and solving the equation:
- Calculate \( f(0) = 0^2 - 6 \times 0 + 4 = 4 \).
This point is essential for graphing as it provides a definitive starting point where the parabola crosses the y-axis.
x-intercepts
The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. These are also known as the roots or zeros of the function. To find these intercepts, set \( f(x) = 0 \) and solve for x using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
In the function \( f(x) = x^2 - 6x + 4 \), the following steps yield the x-intercepts:
These intercepts are essential for sketching the graph as they indicate where the parabola crosses the x-axis.
In the function \( f(x) = x^2 - 6x + 4 \), the following steps yield the x-intercepts:
- Substitute \( a = 1 \), \( b = -6 \), and \( c = 4 \) into the quadratic formula.
- Calculate the discriminant \( b^2 - 4ac = (-6)^2 - 4 \times 1 \times 4 = 20 \).
- Substitute into the formula to get \( x = \frac{6 \pm \sqrt{20}}{2} \).
- Approximate the values to find \( x \approx 1.6 \) and \( x \approx 4.4 \).
These intercepts are essential for sketching the graph as they indicate where the parabola crosses the x-axis.
Graphing Parabolas
Graphing a parabola involves plotting key points and understanding its shape based on the coefficients of the function. The vertex, y-intercept, and x-intercepts provide the necessary points to sketch the parabola accurately.
To graph \( f(x) = x^2 - 6x + 4 \):
To graph \( f(x) = x^2 - 6x + 4 \):
- Begin by plotting the vertex \((3, -5)\), which is the lowest point on the graph for an upward-opening parabola.
- Next, plot the y-intercept \((0, 4)\), a crucial point where the graph crosses the y-axis.
- Then, place the x-intercepts \((1.6, 0)\) and \((4.4, 0)\) on the x-axis, marking where the parabola crosses it.
- Since \( a = 1 \) is positive, indicate that the parabola opens upward, creating a U-shape.
- Draw a smooth curve through these points, ensuring symmetry around the vertex.
