Problem 13 - Basic Technical Mathematics with Calculus

Question

Sketch the graph of each parabola by using the vertex, the \(y\) -intercept, and two other points, not including the \(x\) -intercepts. Check the graph using a calculator. $$y=2 x^{2}+3$$

Step-by-Step Solution

Answer

The vertex is (0, 3), y-intercept is 3, and additional points include (1, 5) and (-1, 5); the graph opens upwards.

Step 1: Identify the Vertex

The equation of the parabola is given in the form \(y = ax^2 + bx + c\). In our case, \(a = 2\), \(b = 0\), and \(c = 3\). The vertex of a parabola in this form is at the point \(h, k\), where \(h = -\frac{b}{2a}\). For this equation, \(h = 0\). Therefore, the vertex point is \((0, 3)\).

Step 2: Find the y-intercept

The y-intercept of a parabola occurs where \(x = 0\). Substitute \(x = 0\) into the equation: \(y = 2(0)^2 + 3 = 3\). Therefore, the y-intercept is at the point \((0, 3)\), which is also the vertex in this case.

Step 3: Find Additional Points

Choose another value for \(x\) and solve for \(y\). For example, if you choose \(x = 1\), substitute into the equation: \(y = 2(1)^2 + 3 = 5\), giving point \((1, 5)\). If you choose \(x = -1\), then \(y = 2(-1)^2 + 3 = 5\), giving point \((-1, 5)\). These points are symmetrical relative to the vertex.

Step 4: Sketch the Graph

Plot the points \((0, 3)\), \((1, 5)\), and \((-1, 5)\) on a graph. Since the parabola is symmetrical about the y-axis, and the vertex is at the minimum point, the graph opens upwards. Connect these points with a smooth curve to form a U-shaped graph.

Step 5: Check Using a Calculator

Input the equation \(y = 2x^2 + 3\) into a graphing calculator. Verify that the graph matches your sketch points with the vertex at \((0, 3)\) and includes points \((1, 5)\) and \((-1, 5)\). Ensure the graph is a symmetric parabola opening upwards.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key point where the curve turns, and it can determine if the parabola opens upwards or downwards. In the standard quadratic form, \(y = ax^2 + bx + c\), the vertex can be found using the formula \(h = -\frac{b}{2a}\) for the x-coordinate, and by substituting back into the equation to find the y-coordinate. By calculation, the vertex for the function \(y = 2x^2 + 3\) is (0, 3). This vertex is crucial because it shows that the parabola touches the y-axis at its highest or lowest turning point in this orientation. Besides, knowing the vertex helps in sketching the basic shape and symmetry of the parabola.

Y-Intercept

The y-intercept is the point where the parabola crosses the y-axis. This occurs when \(x = 0\). For the equation \(y = 2x^2 + 3\), substituting \(x = 0\) results in \(y = 3\). Therefore, the y-intercept is at point (0, 3). This point is valuable when plotting the parabola as it often coincides with the vertex, especially if \(b\) equals zero in the equation. Knowing the y-intercept is like having a touchstone for your graph. Everything else is centered around this key point when dealing with parabola symmetry.

Graphing Calculator

A graphing calculator is a fantastic tool for visualizing the structure of a parabola. By inputting the quadratic equation \(y = 2x^2 + 3\), you can quickly see how the parabola looks. Here's what you'll do:

• Enter the equation into the calculator's graphing mode.
• Check that the parabola opens upwards since the coefficient \(a = 2\) is positive.
• Verify that the vertex (0, 3) is correctly positioned.
• Check that the graph is symmetrical, matching points like (1, 5) and (-1, 5).

A graphing calculator provides you with an immediate visual check of your manual calculations, ensuring accuracy and understanding in your graphing process.

Symmetrical Points

Symmetry is a distinguishing feature of parabolas, and recognizing symmetrical points can simplify graphing. For a parabola shaped like a "U" that opens upwards or downwards, symmetrical points are equidistant from the y-axis on either side of the vertex. In our equation, after identifying the vertex (0, 3), choosing x-values such as 1 and -1 creates points (1, 5) and (-1, 5). These points lie on either side of the y-axis at equal distances from the vertex, confirming symmetry. By plotting these points accurately, you can ensure the curve is drawn smoothly and accurately, revealing the true nature of the parabola's shape.