Problem 33 - A Graphical Approach to Precalculus with Limits

Question

For quadratic function, (a) use the vertex formula to find the coordinates of the vertex and (b) graph the function. Do not use a calculator. $$P(x)=2 x^{2}-2 x+1$$

Step-by-Step Solution

Answer

Vertex is \(\left(\frac{1}{2}, \frac{1}{2}\right)\); graph as an upward-opening parabola.

Step 1: Identify A, B, and C

The given quadratic function is \(P(x) = 2x^2 - 2x + 1\). Identify the coefficients \(A\), \(B\), and \(C\) from the quadratic equation \(Ax^2 + Bx + C\). Here, \(A = 2\), \(B = -2\), and \(C = 1\).

Step 2: Use the Vertex Formula

For a quadratic function \(Ax^2 + Bx + C\), the x-coordinate of the vertex is given by the formula \(x = -\frac{B}{2A}\). Substitute the values of \(B\) and \(A\) to find \(x\):\[x = -\frac{-2}{2 \times 2} = \frac{2}{4} = \frac{1}{2}\]

Step 3: Calculate the Y-coordinate of the Vertex

Substitute \(x = \frac{1}{2}\) back into the function \(P(x) = 2x^2 - 2x + 1\) to find the y-coordinate:\[P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 1\]Calculate each term:\(2 \times \frac{1}{4} = \frac{1}{2}\), \(-2 \times \frac{1}{2} = -1\)Now, combine the terms: \[\frac{1}{2} - 1 + 1 = \frac{1}{2}\]Thus, the y-coordinate is \(\frac{1}{2}\).

Step 4: Determine the Vertex Coordinates

The vertex of the function \(P(x)\) is at the point \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Step 5: Graph the Function

To graph \(P(x) = 2x^2 - 2x + 1\), start by plotting the vertex at \(\left(\frac{1}{2}, \frac{1}{2}\right)\). Since \(A = 2\) is positive, the parabola opens upwards. Next, find additional points by selecting x-values (e.g., 0 or 1) to identify symmetry and shape. Calculate:\[P(0) = 2(0)^2 - 2(0) + 1 = 1\quad \text{(point (0, 1))}\]\[P(1) = 2(1)^2 - 2(1) + 1 = 1\quad \text{(point (1, 1))}\] These points confirm the parabola opens upwards with vertex \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Step 6: Draw the Parabola

Use the vertex and additional points ((0, 1) and (1, 1)) to sketch the parabola. It is symmetrical about the line \(x = \frac{1}{2}\) and passes through the calculated points. Ensure it opens upwards.

Key Concepts

Vertex Formula

The vertex formula is a key component when dealing with quadratic functions. It helps us locate the vertex, or the turning point, of a graph. For any quadratic function of the form \(Ax^2 + Bx + C\), the x-coordinate of the vertex can be found using the formula \(x = -\frac{B}{2A}\). This comes from completing the square or using calculus.

• Identify the coefficients: In the equation \(P(x) = 2x^2 - 2x + 1\), \(A = 2\), \(B = -2\), and \(C = 1\).
• Substitute into the formula to find \(x\): \(-\frac{-2}{2 \times 2} = \frac{1}{2}\).
• Substitute \(x = \frac{1}{2}\) back into the function to find \(y\): \(P(\frac{1}{2}) = \frac{1}{2}\).

Thus, the vertex of this quadratic function is located at \(\ \left(\frac{1}{2}, \frac{1}{2}\right)\). This point is a critical component in understanding and graphing the function.

Graphing Quadratic Functions

Graphing quadratic functions involves several steps, starting with understanding the position and orientation of the graph. For the function \(P(x) = 2x^2 - 2x + 1\), plotting the vertex is the starting point.

• Place the vertex: Begin by graphing the vertex \(\left(\frac{1}{2}, \frac{1}{2}\right)\).
• Determine the direction: Since \(A = 2\) is positive, the parabola opens upwards.
• Find additional points: Select nearby x-values (e.g., \(x = 0\) and \(x = 1\)) and calculate corresponding y-values.

For example, \(P(0) = 1\) yields point (0,1), and \(P(1) = 1\) yields point (1,1). These additional points help confirm the shape of the parabola. The graph will be symmetric about the line \(x = \frac{1}{2}\), which runs through the vertex. Sketch the parabola ensuring its arms extend upwards.

Parabolas

Parabolas are the graph of quadratic functions and have several important properties. A parabola can open upwards or downwards, depending on the sign of the coefficient \(A\) in the quadratic function.

• If \(A > 0\), the parabola opens upwards like a smile.
• If \(A < 0\), it opens downwards like a frown.
• The vertex is the highest or lowest point on the parabola, depending on its direction.

A key feature of parabolas is their symmetry, which is about a vertical line passing through the vertex called the "axis of symmetry". For our function \(P(x) = 2x^2 - 2x + 1\), the axis of symmetry is the line \(x = \frac{1}{2}\). Parabolas appear frequently in different contexts, from physics (projectile motion) to economics (profit graphs). Understanding their properties helps navigate these real-world applications.