Problem 34 - College Algebra

Question

For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. $$ f(x)=x^{2}-2 x $$

Step-by-Step Solution

Answer

Vertex: (1, -1), Axis of symmetry: x = 1, Intercepts: x-intercepts (0, 0) and (2, 0), y-intercept (0, 0).

Step 1: Identify the Form of the Quadratic Function

The given quadratic function is \( f(x) = x^2 - 2x \). This is in the form of \( ax^2 + bx + c \) where \( a = 1 \), \( b = -2 \), and \( c = 0 \).

Step 2: Find the Vertex

The vertex \( (h, k) \) of a quadratic equation in standard form can be found using \( h = -\frac{b}{2a} \). Substituting the values from the given equation, \( h = -\frac{-2}{2(1)} = 1 \). To find \( k \), substitute \( x = 1 \) back into the function: \( f(1) = 1^2 - 2(1) = 1 - 2 = -1 \). Hence, the vertex is \( (1, -1) \).

Step 3: Determine the Axis of Symmetry

The axis of symmetry for the quadratic function can be found using the formula \( x = h \). Since the vertex is \( (1, -1) \), the axis of symmetry is \( x = 1 \).

Step 4: Find the Intercepts

To find the x-intercepts, set \( f(x) = 0 \): \( x^2 - 2x = 0 \). Factor the equation: \( x(x - 2) = 0 \). This gives x-intercepts at \( x = 0 \) and \( x = 2 \). To find the y-intercept, substitute \( x = 0 \) into the function: \( f(0) = 0^2 - 2(0) = 0 \). So, the y-intercept is \( (0, 0) \).

Step 5: Sketch the Graph

Plot the vertex \( (1, -1) \), the axis of symmetry \( x = 1 \), the x-intercepts \( (0, 0) \) and \( (2, 0) \), and the y-intercept \( (0, 0) \) on the graph. Draw a parabolic curve opening upwards that passes through these points, with symmetry about the line \( x = 1 \).

Key Concepts

Vertex

The vertex of a quadratic function is a point where the parabola either reaches its maximum or minimum value. It is essential because it provides a central point around which the parabola is symmetrical. In the standard quadratic equation form, \( ax^2 + bx + c \), finding the vertex involves a couple of straightforward calculations.

The vertex is represented as \((h, k)\). To find \(h\), use the formula \(h = -\frac{b}{2a}\). Here, \(a = 1\) and \(b = -2\), giving us \(h = 1\). Once you have \(h\), calculate \(k\) by substituting \(h\) back into the function: \(f(h)\).

For example:

• Calculate \(f(1) = 1^2 - 2 \times 1 = -1\).

This means the vertex for our function \(f(x) = x^2 - 2x\) is located at \((1, -1)\). This point provides valuable insight into the shape and direction of the parabola.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the graph of a quadratic function into two mirror images, giving the graph its fundamental balanced shape. This axis passes through the vertex of the parabola, and its equation is always \(x = h\), where \(h\) is the x-coordinate of the vertex.

In our particular function, \(f(x) = x^2 - 2x\), we've already calculated \(h\) as 1, thus the axis of symmetry is the line \(x = 1\).

This information becomes useful when sketching or analyzing the quadratic graph because it outlines how the graph behaves on either side of this line. Imagine folding the graph on this line; the two halves would align perfectly, emphasizing the symmetry of a parabola.

Intercepts

Intercepts make it easy to identify where the graph of a quadratic function crosses the axes. This includes both x-intercepts and y-intercepts.

• X-Intercepts: These are the points where the graph crosses the x-axis, meaning the output \(f(x)\) is zero. For the function \(f(x) = x^2 - 2x\), set the equation to zero to solve: \(x^2 - 2x = 0\). By factoring, we get \(x(x - 2) = 0\). Thus, the x-intercepts are \(x = 0\) and \(x = 2\).


• Y-Intercept: This is the point where the graph crosses the y-axis, meaning when \(x = 0\). Substitute \(x = 0\) into the function to find: \(f(0) = 0^2 - 2 \times 0 = 0\). Hence, the y-intercept is at \((0, 0)\).

Knowing the intercepts helps in drawing the initial shape of the quadratic graph and understanding its relationship with the axes. These points help set the frame for sketching the complete curve of the parabola.