Problem 4 - Algebra 1: Concepts and Skills

Question

Sketch the graph of the equation \(y=x^{2}+2 x-4 .\) Plot the point and determine whether it lies inside or outside the parabola. $$ B(-1,3) $$

Step-by-Step Solution

Answer

The vertex of the parabola is at (-1,-5). After plotting point B(-1,3), it can be observed that the point lies outside the parabola.

Step 1: Finding the Vertex

The vertex of a quadratic function in form \(f(x) = ax^{2} + bx + c\) is computed by the formula \((-b/2a, f(-b/2a))\). In this case, \(a = 1, b = 2\), so the vertex is \((-1, f(-1))\). Substitute \(-1\) into the function to find the \(y\)-coordinate of the vertex: \(f(-1) = (-1)^{2} + 2*(-1) - 4 = -1^2 - 2 - 4 = -5\). So, the vertex is \((-1, -5)\).

Step 2: Sketching the Parabola

Now, plot the vertex, and at least two other points on each side of the vertex. Use these points to sketch the curve of the parabola.

Step 3: Determine the Location of the Point

Plot the point B(-1,3) on the graph. This point can be observed visually and determined to be located outside or inside the parabola.

Key Concepts

Vertex of a Parabola

Understanding the vertex of a parabola is crucial when graphing quadratic functions. In simple terms, the vertex is the highest or lowest point on the parabola, depending on whether it opens upwards or downwards. For the quadratic equation of the form \(y = ax^2 + bx + c\), the vertex's \(x\)-coordinate is found using \(\frac{-b}{2a}\), with \(a\) and \(b\) being coefficients from the equation.

To find the \(y\)-coordinate of the vertex, we substitute the \(x\)-coordinate back into the original equation. For example, with the equation \(y = x^2 + 2x - 4\), the vertex is calculated as \(\left(-1, -5\right)\) by first finding the value of \(x\) with \(\frac{-2}{2 \times 1}\) which simplifies to \(\-1\), then substituting \(x = -1\) back into the equation to get the \(y\)-coordinate. This is pivotal in plotting the parabola, as understanding the vertex allows for a more accurate sketch.

It is also important to note that the vertex is a point of symmetry for the parabola, meaning the shape to the left and right of the vertex is a mirror image.

Quadratic Equations

Quadratic equations form the backbone of many algebraic problems and their graphs produce the shape called a parabola. A standard quadratic equation is written as \(y = ax^2 + bx + c\). The \(a\), \(b\), and \(c\) are coefficients that affect the parabola's width, direction, and position on the coordinate plane, respectively.

When \(a > 0\), the parabola opens upwards; when \(a < 0\), it opens downwards. The co-efficient \(b\) affects the horizontal position of the vertex, while \(c\) represents the y-intercept, or where the parabola crosses the \(y\)-axis.

To solve for specific points, we can set \(y\) to different values and solve for \(x\), or vice versa. The solutions to these equations enable us to draw accurate points to help form the parabola. Understanding the effect of each coefficient can help in envisioning the shape even before plotting points.

Parabola Sketching

Creating a rough sketch of a parabola involves plotting points and utilizing the symmetry of the graph about the vertex. After determining the vertex, as illustrated in our example equation \(y = x^2 + 2x - 4\) with vertex \(\left(-1, -5\right)\), plot this central point on a graph. Then, choose additional points on either side of the vertex to ensure the graph is balanced.

The number of points needed to sketch a parabola may vary, but generally, the more points plotted, the more accurate the curve. For clarity and accuracy, selecting points equidistant from the vertex ensures that the plotted graph represents the actual equation. After plotting these points, a curved line is drawn to connect them, forming the parabola.

Finally, to check if a given point lies inside or outside the parabola, plot the point on the same graph. For the given exercise, point \(B(-1,3)\) would be plotted to determine its position relative to the parabola. The position can provide insights into the relationships between the graph and algebraic solutions of the equation.