Problem 187 - Elementary Algebra

Question

In the following exercises, graph by using intercepts, the vertex, and the axis of symmetry. $$ y=2 x^{2}-4 x+1 $$

Step-by-Step Solution

Answer

The vertex is \( (1, -1) \), the axis of symmetry is \( x = 1 \), the x-intercepts are \( 1 \pm \frac{\sqrt{2}}{2} \), and the y-intercept is \( (0, 1) \).

Step 1: Identify the Quadratic Function

The given quadratic function is \( y = 2x^{2} - 4x + 1 \). This is in the standard form \( y = ax^2 + bx + c \). Identify coefficients: \( a = 2 \), \( b = -4 \), and \( c = 1 \).

Step 2: Find the x-intercepts

To find the x-intercepts, set \( y = 0 \) and solve for \( x \):\( 0 = 2x^2 - 4x + 1 \).Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = 1 \pm \frac{\sqrt{2}}{2} \].Thus, the x-intercepts are \( x = 1 + \frac{\sqrt{2}}{2} \) and \( x = 1 - \frac{\sqrt{2}}{2} \).

Step 3: Find the y-intercept

To find the y-intercept, set \( x = 0 \):\( y = 2(0)^2 - 4(0) + 1 \) which simplifies to \( y = 1 \).Thus, the y-intercept is \( (0, 1) \).

Step 4: Find the Vertex

The vertex of a parabola given by \( y = ax^2 + bx + c \) can be found using the formula \( x = \frac{-b}{2a} \):\( x = \frac{-(-4)}{2 \cdot 2} = \frac{4}{4} = 1 \).Plug \( x = 1 \) back into the equation to get the corresponding \( y \):\( y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = -1 \).Thus, the vertex is \( (1, -1) \).

Step 5: Determine the Axis of Symmetry

The axis of symmetry of a parabola in the form \( y = ax^2 + bx + c \) is given by \( x = \frac{-b}{2a} \).Using \( b = -4 \) and \( a = 2 \), we have:\( x = \frac{4}{4} = 1 \).Thus, the axis of symmetry is the vertical line \( x = 1 \).

Step 6: Graphing the Function

Plot the x-intercepts \( (1 + \frac{\sqrt{2}}{2}, 0) \) and \( (1 - \frac{\sqrt{2}}{2}, 0) \), the y-intercept \( (0, 1) \), and the vertex \( (1, -1) \). Draw the axis of symmetry as the vertical line \( x = 1 \). Finally, sketch the parabolic curve that passes through these points, opening upwards.

Key Concepts

Intercepts in Quadratic Functions

Intercepts are crucial points on a graph where the curve crosses the axes. For quadratic functions, we focus on both x-intercepts and y-intercepts.

To find the x-intercepts, we need to set the equation to zero and solve for x. As shown in the exercise, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
This formula gives us the solutions where the function intersects the x-axis.

For our example, the quadratic equation was solved, and the x-intercepts were found to be:
\[ x = 1 \pm \frac{\sqrt{2}}{2} \].
Please note, the \( \pm \) symbol means we have two x-intercepts. When plotted on the graph, these points show exactly where the parabola crosses the x-axis.

The y-intercept is easier to find. Set \( x = 0 \) in the equation and solve for y. For our function, \( y = 2(0)^{2} - 4(0) + 1 \), which simplifies to \( y = 1 \). So, the y-intercept is at \( (0, 1) \). This point shows where the parabola crosses the y-axis.

Vertex of a Parabola

The vertex is a key point on a parabola and represents its highest or lowest point depending on whether it opens upwards or downwards.

To find the vertex of a quadratic function in the form \( y=ax^2+bx+c \), we use the formula:
\[ x = \frac{-b}{2a} \].
In the exercise, with \( a = 2 \) and \( b = -4 \), we found:
\[ x = \frac{-(-4)}{2 \cdot 2} = \frac{4}{4} = 1 \].
Then, plug \( x = 1 \) back into the function to find the y-coordinate:
\[ y = 2(1)^2 - 4(1) + 1 = -1 \].
Thus, the vertex is \( (1, -1) \).

This point is central as it represents the minimum point of our parabola (since it opens upwards) and helps in defining the shape and position of the graph.
When graphing, placing the vertex accurately ensures the correct depiction of the curve's behavior.

Axis of Symmetry in Parabolas

The axis of symmetry is an imaginary vertical line that divides the parabola into two mirror-image halves. Every quadratic function has an axis of symmetry, and it passes through the vertex.

For a quadratic function in the form \( y = ax^2 + bx + c \), the formula to find the axis of symmetry is:
\[ x = \frac{-b}{2a} \].
In our exercise, with \( b = -4 \) and \( a = 2 \), we have:
\[ x = \frac{-(-4)}{2 \cdot 2} = 1 \].
Therefore, the axis of symmetry is the vertical line at \( x = 1 \).

This line is crucial because it ensures precision while drawing the parabola. By knowing where the axis of symmetry lies, we can accurately reflect the shape and form of the graph. Always plot this line first, immediately after plotting the vertex, for better accuracy in graphing.