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Magazine Chapter 3: Problem 14
A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x\) - and \(y\) -intercept(s). (c) Sketch its graph. $$f(x)=x^{2}-2 x+2$$
Short Answer
Expert verified
The vertex is \((1, 1)\), no x-intercepts, y-intercept is \((0, 2)\).
Step by step solution
01
Identify the Quadratic Function
The given quadratic function is \( f(x) = x^2 - 2x + 2 \). This function is already in the form \( ax^2 + bx + c \), where \( a = 1 \), \( b = -2 \), and \( c = 2 \).
02
Express in Standard Form
To express the function in standard form, which is \( f(x) = a(x-h)^2 + k \), complete the square. Start with \( f(x) = x^2 - 2x + 2 \), and write it as \( x^2 - 2x \). The term to complete the square is \( (b/2)^2 = (-2/2)^2 = 1 \). Thus: \[ f(x) = (x^2 - 2x + 1) - 1 + 2 \]which simplifies to:\[ f(x) = (x-1)^2 + 1 \].
03
Find the Vertex
The vertex of the parabola in the form \( f(x) = a(x-h)^2 + k \) is given by \( (h, k) \). From \( f(x) = (x-1)^2 + 1 \), we have \( h = 1 \) and \( k = 1 \). Thus, the vertex is \( (1, 1) \).
04
Find the x-intercepts
To find the x-intercepts, set \( f(x) = 0 \) and solve for \( x \):\[ (x-1)^2 + 1 = 0 \]\[ (x-1)^2 = -1 \].No real solutions exist because a square cannot equal a negative number. Hence, there are no x-intercepts.
05
Find the y-intercept
The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the original equation:\[ f(0) = 0^2 - 2\times0 + 2 = 2 \].Thus, the y-intercept is \( (0, 2) \).
06
Sketch the Graph
Since the vertex is at \( (1, 1) \) and the parabola opens upwards (because the coefficient \( a = 1 \) is positive), sketch a graph with the vertex at \( (1, 1) \) and passing through \( (0, 2) \). The curve doesn't cross the x-axis, reflecting having no x-intercepts.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex Form
In quadratic functions, the vertex form is a way to express the equation to highlight the vertex and the direction of the parabola. The general vertex form is given by \( f(x) = a(x-h)^2 + k \). Here:
- \( a \) indicates the direction and the width of the parabola. - \( (h, k) \) is the vertex of the parabola, giving you a precise point through which the function passes and pivots.
Expressing a function in this form involves completing the square. For example, with \( f(x) = x^2 - 2x + 2 \), the vertex form is derived to \( f(x) = (x-1)^2 + 1 \). This shows the vertex at \( (1,1) \) and lets you know right away that the parabola's minimum point is at this vertex.
- \( a \) indicates the direction and the width of the parabola. - \( (h, k) \) is the vertex of the parabola, giving you a precise point through which the function passes and pivots.
Expressing a function in this form involves completing the square. For example, with \( f(x) = x^2 - 2x + 2 \), the vertex form is derived to \( f(x) = (x-1)^2 + 1 \). This shows the vertex at \( (1,1) \) and lets you know right away that the parabola's minimum point is at this vertex.
Vertex
The vertex of a quadratic function is a critical point that represents either the peak or the trough of the parabola. It's the point \((h, k)\) in the vertex form \( f(x) = a(x-h)^2 + k \). For the function \( f(x) = (x-1)^2 + 1 \), the vertex is \( (1, 1) \).
Understanding the vertex helps us determine:
Understanding the vertex helps us determine:
- The symmetry of the parabola, which mirrors across the line \( x = h \).
- Whether the vertex is the highest or lowest point, based on \( a \). If \( a > 0 \), it's the minimum point and if \( a < 0 \), it's the maximum point.
X-intercept
An x-intercept is where the parabola meets the x-axis, corresponding to an \( y \)-value of zero. In mathematics, finding the x-intercepts involves solving \( f(x) = 0 \) for \( x \). Starting from \( (x-1)^2 + 1 = 0 \), we find there are no real solutions since \( (x-1)^2 = -1 \). Since a squared term cannot be negative, the graph does not cross the x-axis.
When a quadratic function lacks x-intercepts, it suggests the vertex is either entirely above or below the x-axis, depending on the parabola's orientation.
When a quadratic function lacks x-intercepts, it suggests the vertex is either entirely above or below the x-axis, depending on the parabola's orientation.
Y-intercept
To find the y-intercept, evaluate the function at \( x = 0 \). This tells you where the parabola crosses the y-axis. For \( f(x) = x^2 - 2x + 2 \), substituting \( x = 0 \) gives: \[ f(0) = 0^2 - 2\cdot0 + 2 = 2 \] Thus, the y-intercept is \( (0, 2) \).
The y-intercept is useful as another anchor point to sketch the graph, along with the vertex.
It offers a starting point to understand the range and direction of the parabola as it progresses vertically from the y-axis.
The y-intercept is useful as another anchor point to sketch the graph, along with the vertex.
It offers a starting point to understand the range and direction of the parabola as it progresses vertically from the y-axis.
Parabola Graphing
Graphing a parabola involves marking key points such as the vertex and intercepts and understanding the parabola's orientation based on the \( a \) coefficient. With \( f(x) = (x-1)^2 + 1 \), the steps for graphing would be:
These actions collectively give a complete picture of how the parabola behaves in its domain.
- Plot the vertex \( (1, 1) \). This is the pivot point.
- Identify the y-intercept at \( (0, 2) \).
- As \( a = 1 \) (positive), plot the parabola opening upwards with symmetrical arms equidistant from the vertex line \( x = 1 \).
- In the absence of x-intercepts, ensure the parabola doesn鈥檛 touch the x-axis.
These actions collectively give a complete picture of how the parabola behaves in its domain.
