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Magazine Chapter 3: Problem 47
Identify the vertex, axis of symmetry, y-intercept, x-intercepts, and opening of each parabola, then sketch the graph. $$f(x)=(x-3)^{2}-4$$
Short Answer
Expert verified
Vertex: (3, -4); Axis of symmetry: x = 3; Y-intercept: (0, 5); X-intercepts: (1, 0) and (5, 0); Opens upwards.
Step by step solution
01
Identify the vertex
The vertex form of a quadratic function is given by \[ f(x) = a(x-h)^2 + k \] where (h, k) is the vertex. For the function \[ f(x) = (x-3)^2 - 4 \], the vertex is (3, -4).
02
Find the axis of symmetry
The axis of symmetry of a parabola is the vertical line that passes through the vertex. For the function \[ f(x) = (x-3)^2 - 4 \], the axis of symmetry is \[ x = 3 \].
03
Determine the y-intercept
The y-intercept is the point where the parabola crosses the y-axis. To find the y-intercept, set x=0 and solve for f(x): \[ f(0) = (0-3)^2 - 4 = 9 - 4 = 5 \]. So the y-intercept is (0, 5).
04
Calculate the x-intercepts
The x-intercepts are the points where the parabola crosses the x-axis. Set f(x) = 0 and solve for x: \[ 0 = (x-3)^2 - 4 \] Add 4 to both sides: \[ 4 = (x-3)^2 \] Take the square root of both sides: \[ \text{卤}2 = x-3 \] Solve for x: \[ x = 5 \text{ or } x = 1 \] So the x-intercepts are (5, 0) and (1, 0).
05
Determine the opening of the parabola
The coefficient of the term \((x-3)^2\) indicates the direction of the parabola's opening. Since the coefficient is positive, the parabola opens upwards.
06
Sketch the graph
Plot the vertex (3, -4), the y-intercept (0, 5), and the x-intercepts (1, 0) and (5, 0). Draw the parabola opening upwards with the axis of symmetry as the vertical line x = 3.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Form
The vertex form of a quadratic function is a specific way of writing the function to easily identify its key features. It is written as: \[ f(x) = a(x-h)^2 + k \] In this form, the constants \( h \) and \( k \) reveal the vertex \( (h, k) \) . For example, in the function \( f(x) = (x-3)^2 - 4 \) , comparing it to the vertex form, we can see that \( h = 3 \) and \( k = -4 \). Thus, the vertex is (3, -4). The vertex form makes it straightforward to find the vertex, providing a central point from which the quadratic function unfolds.
Axis of Symmetry
The axis of symmetry is a vertical line that runs through the vertex of the parabola. This line divides the parabola into two mirror-image halves. For any quadratic function \[ f(x) = a(x-h)^2 + k \] , the axis of symmetry can be found using the formula \( x = h \). In our example, for the function \[ f(x) = (x-3)^2 - 4 \] , the axis of symmetry is at \( x = 3 \). This means that if you fold the graph along the line \( x = 3 \), both halves of the parabola would match perfectly, showing the symmetrical nature of quadratic functions.
Y-Intercept
The y-intercept is the point where the graph of the quadratic function crosses the y-axis. To find the y-intercept in any quadratic function, substitute \( x = 0 \) and solve for \( f(x) \). For the function \[ f(x) = (x-3)^2 - 4 \] , we set \( x = 0 \): \[ f(0) = (0-3)^2 - 4 = 9 - 4 = 5 \] . Therefore, the y-intercept is \( (0, 5) \). This point indicates where the parabola touches the y-axis, providing a helpful reference for graphing the function.
X-Intercepts
X-intercepts are the points where the graph crosses the x-axis (i.e., where \( f(x) = 0 \)). To find the x-intercepts of a quadratic function, set the function equal to zero and solve for \( x \). For \[ f(x) = (x-3)^2 - 4 \] , we solve: \[ 0 = (x-3)^2 - 4 \] . Adding 4 to both sides: \[ 4 = (x-3)^2 \] . Taking the square root of both sides gives: \[ \text{卤}2 = x - 3 \] . Solving for \( x \) yields: \[ x = 5 \text{ and } x = 1 \] . Thus, the x-intercepts are at \( (5, 0) \) and \( (1, 0) \). These points indicate where the parabola intersects the x-axis, and are also known as the roots of the quadratic equation.
Parabola Opening
The direction in which a parabola opens is determined by the coefficient of the \( (x-h)^2 \) term in the vertex form of the quadratic function. If this coefficient \( a \) is positive, the parabola opens upwards, like a U-shape. If \( a \) is negative, it opens downwards, like an upside-down U. In our example \[ f(x) = (x-3)^2 - 4 \] , the coefficient of \( (x-3)^2 \) is 1, which is positive. Therefore, the parabola opens upwards. This piece of information is crucial when sketching the graph, as it determines the general shape and direction the parabola will take.
