Problem 92 - College Algebra

Question

Graph each parabola with the given equation. \(y=x^{2}+4 x-5\)

Step-by-Step Solution

Answer

The graph of the equation \(y=x^{2}+4x-5\) is a parabola with vertex (-2,-5) and x-intercepts (1,0) and (-5,0). The parabola opens upwards as \(a>0\).

Step 1: Identify the Vertex of the Parabola

The first step is to identify the vertex of the parabola. This is in the form \(h, k\), where \(h= -b/2a\), and 'k' is the y value when 'x=h'. For the equation \(y=x^{2}+4 x-5\), \(a=1, b=4\) and \(c=5\). So, \(h=-b/2a=(-4)/(2*1)=-2\). Substituting \(x=-2\) into the equation, we get \(y=(-2)^{2}+4(-2)-5=-5\). So the vertex of the parabola is (-2,-5).

Step 2: Find the X intercepts

The next step is finding the x-intercepts. To do this, set \(y=0\) and solve for 'x'. So, \(0=x^{2}+4 x-5\). By factorization method, we get \(0=(x-1)(x+5)\). Setting each bracket to zero, we get \(x=1\) and \(x=-5\). So, the x-intercepts of the parabola are '(1,0)' and '(-5,0)'.

Step 3: Plot the Vertex and X intercepts on a Graph

Plot the vertex '(-2,-5)', and the x intercepts '(1,0)' and '(-5,0)'. Use these points to draw a rough sketch of the parabola. Remember that the line of symmetry of the parabola is the vertical line \(x=h\), so the x intercepts are symmetric about the line \(x=h\).

Step 4: Draw the Parabola

Now, join these points to draw a smooth curve making sure the plotted points are located symmetrically about the line of symmetry \(x=-2\). This will ensure that the parabola opens upwards as \(a=1\) which is greater than 0.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is the point where it changes direction. It's a crucial point that helps in understanding the parabola's position and shape. For any quadratic equation given in the form \(y = ax^2 + bx + c\), the vertex is situated at \((h, k)\). Here, \(h = -\frac{b}{2a}\), and \(k\) is the y-value when \(x = h\). To find \(k\), substitute \(x = h\) back into the equation. In our example, with the equation \(y = x^2 + 4x - 5\), the vertex is \((-2, -5)\). Here's how we got it:

• Calculate \(h\): \(-\frac{4}{2 \cdot 1} = -2\).
• Find \(k\) by substituting into the equation: \(( -2)^2 + 4(-2) - 5 = -5\).

Understanding the vertex helps in graphing the parabola and determining where it reaches its maximum or minimum point.

X-Intercepts

The x-intercepts of a parabola are the points where the graph crosses the x-axis. They are significant as they give an idea of the horizontal position of the parabola. To find the x-intercepts, set the quadratic equation equal to zero and solve for \(x\).
For our equation, \(y = x^2 + 4x - 5\), do the following:

• Set \(y=0\) to get: \(0 = x^2 + 4x - 5\).
• Factor the equation: \(0 = (x - 1)(x + 5)\).
• Solve each factor equation: \(x - 1 = 0\) or \(x + 5 = 0\), leading to \(x = 1\) and \(x = -5\).

Hence, the x-intercepts are \((1, 0)\) and \((-5, 0)\), telling us where the parabola touches the x-axis.

Line of Symmetry

The line of symmetry of a parabola is an imaginary vertical line that passes through the vertex and divides the parabola into two mirror-image halves. It helps to ensure that the parabola is symmetric and provides a reference point for sketching.
For any quadratic equation \(y = ax^2 + bx + c\), the line of symmetry can be identified by the equation \(x = h\), where \(h\) is the x-coordinate of the vertex. In the equation \(y = x^2 + 4x - 5\), the line of symmetry is \(x = -2\).
By knowing the line of symmetry, we can use it to position points accurately on a graph, ensuring the parabola is drawn correctly. Each point on one side of this line has a corresponding point on the opposite side.

Sketching Parabolas

Sketching a parabola involves combining all the elements: the vertex, x-intercepts, line of symmetry, and the general shape. Begin by plotting these essential points on a coordinate plane.

• Start with the vertex, which is \((-2, -5)\) for our equation.
• Next, plot the x-intercepts: \((1, 0)\) and \((-5, 0)\).
• Draw the line of symmetry \(x = -2\) as a guideline for symmetry.
• Finally, connect these points with a smooth, upward-opening curve, because the coefficient \(a\) is positive, indicating the parabola opens upwards.

This method ensures a precise sketch of the parabola, capturing all key features and visually representing the quadratic equation.