Problem 12 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(y=(x+1)^{2}-5\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(y = (x + 1)^2 - 5\) is at \((-1, -5)\). The axis of symmetry is given by the line \(x = -1\). The x-intercepts are \((-1 + \sqrt{5}, 0)\) and \((-1 - \sqrt{5}, 0)\), and the y-intercept is at \((0, -4)\). To graph the function, plot the vertex, axis of symmetry, x-intercepts, and y-intercept, and sketch a parabola connecting these points.

Step 1: Find the vertex

To find the vertex of the given quadratic function, first, we need to rewrite it in vertex form: \(y = a(x - h)^2 + k\) Where (h, k) is the vertex of the parabola. We can see that our function is already in this form: \(y = (x + 1)^2 - 5\) So, the vertex (h, k) is (-1, -5).

Step 2: Find the axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex. Its equation is given by: \(x = h\) Since the vertex is (-1, -5), the axis of symmetry is: \(x = -1\)

Step 3: Find the x-intercepts

To find the x-intercepts, we need to set y = 0 and solve for x: \((x + 1)^2 - 5 = 0\) Now solve for x: 1. \((x + 1)^2 = 5\) 2. \(x + 1 = \pm\sqrt{5}\) 3. \(x = -1 \pm\sqrt{5}\) So, the x-intercepts are \((-1 + \sqrt{5}, 0)\) and \((-1 - \sqrt{5}, 0)\).

Step 4: Find the y-intercept

To find the y-intercept, we need to set x = 0 and solve for y: \(y = (0 + 1)^2 - 5\) Simplify: \(y = 1 - 5\) \(y = -4\) So, the y-intercept is \((0, -4)\).

Step 5: Graph the function

To graph the function, follow these steps: 1. Plot the vertex at (-1, -5). 2. Draw the axis of symmetry at x = -1. 3. Plot the x-intercepts at \((-1 + \sqrt{5}, 0)\) and \((-1 - \sqrt{5}, 0)\). 4. Plot the y-intercept at (0, -4). 5. Sketch a parabola by connecting these points. By following these steps, you will have a graph of the quadratic function \(y = (x + 1)^2 - 5\).

Key Concepts

Vertex

In any quadratic function of the form \(y = a(x - h)^2 + k\), the vertex is the point \((h, k)\). This point is the peak or the lowest point on the graph, depending on whether the parabola opens upwards or downwards. In the quadratic function given, \(y = (x+1)^2 - 5\), it is evident that the function is already in the vertex form. This makes identifying the vertex straightforward: \(h = -1\) and \(k = -5\), so the vertex is at \((-1, -5)\).

The vertex is a crucial point on the graph because it represents either the maximum or minimum value of the quadratic function. Since quadratic functions create parabolas, for functions of the form \(a(x - h)^2 + k\) with \(a > 0\) (as in this case where \(a = 1\)), the parabola opens upwards, and the vertex is the minimum point.

Axis of Symmetry

The axis of symmetry is an invisible vertical line that divides the parabola into two mirror-image halves. Its equation can be derived directly from the vertex's x-coordinate. For any function \(y = a(x - h)^2 + k\), the axis of symmetry is \(x = h\).

For our quadratic function \(y = (x+1)^2 - 5\), since the vertex is at \((-1, -5)\), the axis of symmetry is \(x = -1\).
This line not only helps in drawing the parabola correctly but also ensures that symmetric points across this line have the same y-value.

• The axis of symmetry always passes through the vertex.
• When graphed, it serves as a guide for ensuring the symmetry of the parabola.

X-Intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. These are found by setting \(y = 0\) and solving the resulting equation for \(x\).

In our equation \(y = (x+1)^2 - 5\), set \(y = 0\):

• \((x+1)^2 - 5 = 0\)
• \((x+1)^2 = 5\)
• \(x+1 = \pm \sqrt{5}\)
• \(x = -1 \pm \sqrt{5}\)

This tells us the x-intercepts are at \((-1 + \sqrt{5}, 0)\) and \((-1 - \sqrt{5}, 0)\).

Finding x-intercepts is important in understanding the roots (solutions) of the quadratic equation and helps in verifying and graphing the quadratic function accurately.

Y-Intercepts

The y-intercept is where the graph of the function crosses the y-axis. It can be found by substituting \(x = 0\) into the equation and solving for \(y\).
For the function \(y = (x+1)^2 - 5\), when \(x = 0\):

• \(y = (0+1)^2 - 5\)
• \(y = 1 - 5\)
• \(y = -4\)

Thus, the y-intercept is at \((0, -4)\).

The y-intercept provides a starting point for the graph on the y-axis and is essential for sketching the function as it helps determine the initial direction and steepness of the parabola from the y-axis.