Problem 28 - Introductory Algebra for College Students

Question

Graph the parabola whose equation is given $$y=x^{2}+4 x-5$$

Step-by-Step Solution

Answer

The parabola of the given equation has its vertex at (-2, -9) and crosses the y-axis at (0, -5). It opens upwards, with the axis of symmetry being at x=-2.

Step 1: Determine the Vertex

The vertex of a parabola given in the form \(y=a(x-h)^{2}+k\) has the coordinates (h, k). Applying this to our equation, we first put it into vertex form by completing the square: \(y=(x^{2}+4x+4)-4-5 =(x+2)^{2}-9\). This gives the vertex at (-2, -9). This is the point where the parabola either has a minimum or maximum value.

Step 2: Determine the Y-Intercept

The y-intercept is the point where the parabola crosses the y-axis (i.e., where x=0). Substituting x=0 into the equation gives \(y=0^{2}+4*0-5=-5\). Therefore, the y-intercept is at (0,-5).

Step 3: Plot the Vertex and the Y-Intercept

Begin plotting the parabola by marking the vertex at (-2,-9) and the y-intercept at (0, -5). The axis of symmetry is the line x=h, which in this case is x=-2. The parabola opens upwards because the coefficient of \(x^{2}\) is positive.

Step 4: Sketch the Parabola

Finally, sketch the parabola making sure that it passes through the points plotted and that it is symmetric about the x=-2 line. As it opens upwards, the vertex will be the minimum point of the parabola.

Key Concepts

Vertex Form

To graph a parabola efficiently, it's helpful to use its vertex form, which is expressed as \(y = a(x - h)^2 + k\). This form clearly shows the vertex of the parabola at point \((h, k)\). Understanding vertex form allows you to easily identify key features of the parabola, such as the vertex, which can be either a minimum or maximum point, depending on the parabola's orientation.

To convert the standard form of a quadratic equation \(y = ax^2 + bx + c\) into vertex form, we use a technique called completing the square, which can seem intimidating at first, but is actually quite logical and systematic.

Completing the Square

Completing the square is a technique used to convert the standard quadratic equation into vertex form. This step is crucial for finding the vertex without the need for calculus. Let's break down the process:

• Start with the equation \(y = x^2 + 4x - 5\).
• Focus on the \(x^2\) and \(x\) terms: \(x^2 + 4x\).
• Add and subtract the square of half the coefficient of \(x\) (which is 2 in this case), giving us \((x^2 + 4x + 4) - 4\).
• Factor \((x^2 + 4x + 4)\) to get \((x + 2)^2\).
• Rewrite the original equation including the constant term, resulting in \((x + 2)^2 - 9\).

This manipulation transforms the equation into vertex form, \(y = (x + 2)^2 - 9\), making it much easier to identify the vertex.

Y-Intercept

The y-intercept of a parabola is the point where it crosses the y-axis. In simpler terms, it's the value of \(y\) when \(x = 0\). To determine the y-intercept from the equation \(y = x^2 + 4x - 5\), substitute \(x = 0\) and solve for \(y\).

• Substitute \(x = 0\) into the equation: \(y = 0^2 + 4(0) - 5 = -5\).
• Thus, the y-intercept is the point \((0, -5)\).

The y-intercept is crucial for graphing as it provides a clear point where the graph will cross the y-axis, aiding in the overall shape and orientation of the parabola on the coordinate plane.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For any parabola, the axis of symmetry goes through the vertex, and its equation is \(x = h\). In vertex form \(y = a(x - h)^2 + k\), \(h\) represents the x-coordinate of the vertex.

For our example, having transformed the equation to \((x + 2)^2 - 9\) means that the vertex is at \((-2, -9)\), making the axis of symmetry \(x = -2\). This concept is important because it will guide you in accurately sketching the parabola, ensuring each side mirrors the other perfectly.

Understanding the axis of symmetry helps not only in plotting just half the points needed to draw the parabola but also confirms that your graph is correctly centered around \(x = -2\). Keeping this symmetry in mind is critical for an accurate representation.