Problem 5 - Algebra 2

Question

Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=2 x^{2}-4 x+1 $$

Step-by-Step Solution

Answer

The y-intercept is 1, the axis of symmetry is \(x=1\), and the vertex is \((1, -1)\). Graph using points \((0, 1)\), \((1, -1)\), and \((2, 1)\).

Step 1: Find the Y-Intercept

To find the y-intercept, substitute \(x = 0\) into the function \(f(x) = 2x^2 - 4x + 1\). This gives \(f(0) = 2(0)^2 - 4(0) + 1 = 1\). Therefore, the y-intercept is 1.

Step 2: Find the Axis of Symmetry and X-Coordinate of the Vertex

The axis of symmetry for a quadratic function \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). For the function \(f(x) = 2x^2 - 4x + 1\), here \(a = 2\) and \(b = -4\). Substitute these values into the formula to find the axis of symmetry: \(x = -\frac{-4}{2(2)} = 1\). Thus, the axis of symmetry is \(x = 1\), which is also the x-coordinate of the vertex.

Step 3: Compute the Vertex Y-Coordinate

With the x-coordinate of the vertex found from the axis of symmetry as \(x = 1\), substitute it back into the function to find the y-coordinate of the vertex: \(f(1) = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1\). So, the vertex is at \((1, -1)\).

Step 4: Make a Table of Values

Constructing a table of values around the vertex \((1, -1)\), choose values of \(x\) such as 0, 1, and 2. Calculate \(f(x)\) for each of these values:1. \(x = 0\), \(f(0) = 1\)2. \(x = 1\), \(f(1) = -1\) (vertex)3. \(x = 2\), \(f(2) = 2(2)^2 - 4(2) + 1 = 1\)The table is:\[\begin{array}{c|c}x & f(x) \\hline0 & 1 \1 & -1 \2 & 1 \\end{array}\]

Step 5: Graph the Function

Plot the points from the table: \((0, 1)\), \((1, -1)\), and \((2, 1)\). Draw a smooth parabola that passes through these points, ensuring that the vertex \((1, -1)\) is the lowest point (since the parabola opens upward, as indicated by the positive coefficient of \(x^2\)) and that the axis of symmetry is \(x = 1\).

Key Concepts

Vertex Form of a Quadratic

Quadratic functions can be expressed in different forms, with one of the most helpful being the vertex form. The vertex form of a quadratic equation is written as \( f(x) = a(x - h)^2 + k \). In this format, \((h, k)\) represents the vertex of the parabola.

• The value \( h \) shows the \( x \)-coordinate of the vertex, telling us exactly where the parabola turns on the \( x \)-axis.
• The value \( k \) provides the \( y \)-coordinate of the vertex, revealing how high or low the parabola opens.

Understanding the vertex form simplifies the process of finding and plotting the vertex, which is crucial for graph sketching. For the given quadratic function \( f(x) = 2x^2 - 4x + 1 \), the standard form can be converted to vertex form to offer insights about the function.

Axis of Symmetry

The axis of symmetry of a quadratic function is a vertical line that divides the parabola into two mirror-image halves. Every point on one side of this axis is mirrored by a point on the other side.For a function in the form \( ax^2 + bx + c \), the axis of symmetry can be calculated using the formula:\[ x = -\frac{b}{2a} \]In the example of \( f(x) = 2x^2 - 4x + 1 \), by substituting the values \( b = -4 \) and \( a = 2 \), we find the axis of symmetry is \( x = 1 \). This means the parabola's "mirror line" is vertically aligned at \( x = 1 \), and it is also the \( x \)-coordinate of the vertex. Recognizing this axis helps in graphing the quadratic accurately and understanding its alignment.

Y-Intercept

The y-intercept of a quadratic function is the point where the graph crosses the \( y\)-axis. This is the point where \( x = 0 \). To find the y-intercept, substitute \( x = 0 \) into the function and solve for \( f(x) \).For the function \( f(x) = 2x^2 - 4x + 1 \):\[ f(0) = 2(0)^2 - 4(0) + 1 = 1 \]This calculation shows that the y-intercept is at the point \( (0, 1) \).- The y-intercept is useful as a starting point for plotting the graph.- It also gives immediate information about the graph's behavior when \( x \) is zero.With the y-intercept in hand, along with the vertex and axis of symmetry, creating a complete sketch of the function becomes manageable and straightforward.