Problem 15 - 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 \(y=-(x+1)^{2}-5\) is at (-1, -5), and the axis of symmetry is \(x=-1\). There are no x-intercepts, and the y-intercept is at the point (0, -6). The graph of the function is a parabola opening downwards, with the vertex, axis of symmetry, and y-intercept plotted accordingly.

Step 1: Identify the vertex

Since the given function is \(y=-(x+1)^{2}-5\), we can compare it with the vertex form \(y=a(x-h)^{2}+k\). We have \(a=-1\), \(h=-1\), and \(k=-5\). Therefore, the vertex of the function is at \((h,k)=(-1,-5)\).

Step 2: Identify the axis of symmetry

The axis of symmetry is a vertical line passing through the vertex. Since the vertex has an x-coordinate of -1, the equation for 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: \(0=-(x+1)^{2}-5\) Add 5 to both sides: \(5=-(x+1)^{2}\) Now, divide both sides by -1: \(-5=(x+1)^{2}\) Since we cannot square a real number to get a negative value, there are no x-intercepts for this function.

Step 4: Find the y-intercept

To find the y-intercept, we need to set x = 0 and solve for y: \(y=-(0+1)^{2}-5\) \(y=-1^{2}-5\) \(y=-1-5\) \(y=-6\) The y-intercept is at the point (0, -6).

Step 5: Graph the function

Now that we have identified the vertex, axis of symmetry, and y-intercept, we can graph the function. Plot the vertex at (-1, -5), and the y-intercept at (0, -6). Since there are no x-intercepts, the graph will open downwards and not touch the x-axis. Draw the axis of symmetry at \(x=-1\). Sketch the parabola symmetrically on either side of the axis of symmetry, as the parabola opens downwards due to \(a=-1\).

Key Concepts

Vertex Form

In quadratic functions, the vertex form is a convenient way to express the equation, especially when identifying key features like the vertex. The vertex form is given by the equation:\[ y = a(x-h)^2 + k \]This format helps in easily determining the vertex \(h, k\). For the function \( y = -(x+1)^2 - 5 \), by comparing it with the vertex form, we can identify \( a = -1 \), \( h = -1 \), and \( k = -5 \). Therefore, the vertex is at the point \(-1, -5\). The vertex is crucial as it gives us a point on the graph and helps in sketching the parabola. It is the maximum or minimum point of the function, depending on whether the parabola opens upwards or downwards.

Axis of Symmetry

The axis of symmetry is an important characteristic of any quadratic function. It is a vertical line that passes through the vertex, dividing the parabola into two mirror images. The equation for the axis of symmetry can be extracted directly from the vertex form equation. It is given by:\[ x = h \]Where \( h \) is the x-coordinate of the vertex. For our function \( y = -(x+1)^2 - 5 \), we determined the vertex to be \(-1, -5\), hence the axis of symmetry is \( x = -1 \). This symmetry helps in graphing because once you plot the vertex and know the direction of the opening, you can mirror points across this axis for precise parabola graphing.

Parabola Graphing

Graphing a parabola involves understanding its basic features such as the vertex, axis of symmetry, and how it opens. For the quadratic \( y = -(x+1)^2 - 5 \), we already know:

• Vertex: \(-1, -5\)
• Axis of Symmetry: \( x = -1 \)

To draw the graph:

• Plot the vertex at \(-1, -5\).
• Draw the axis of symmetry, a dashed vertical line at \( x = -1 \).
• Since \( a = -1 \), indicating the parabola opens downwards, sketch the curve opening downward.
• No x-intercepts present, as the function doesn't touch the x-axis.

By using the axis of symmetry, you can ensure the parabola is symmetric around this line, providing a mirror image on both sides.

Y-intercept Calculation

The y-intercept of a function is the point where the graph crosses the y-axis. To find the y-intercept, set \( x = 0 \) in the quadratic equation and solve for \( y \). For our function:\[ y = -(0+1)^2 - 5 \]Calculating the expression:

• First, substitute 0 for x, giving us \( -1^2 \).
• Compute \( -1^2 = -1 \).
• Then subtract 5: \( -1 - 5 = -6 \).

Therefore, the y-intercept is located at the point \( (0, -6) \). This point is critical when plotting the graph, as it gives an additional reference point to confirm the parabola's position and orientation. Remember, to properly sketch, consider the parabola's symmetry, direction of opening, and these intercept calculations.