Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=(x+5)^{2}-6$$

Short answer

The vertex of the function is at \(-5, -6\), the y-intercept is at (0,19), and the x-intercepts are at \(\sqrt{6} - 5\) and \(-\sqrt{6} - 5\). The function graph is a parabola that opens upwards.

Step-by-step solution

Write the function in vertex form

The given function is already in vertex form \(f(x) = a(x-h)^{2} + k\). So, the coordinates of the vertex can be read directly from the equation: \(h = -5\) and \(k = -6\). Thus, the vertex of the function is \(-5, -6\).

Find the intercepts

The x-intercepts can be found by setting \(f(x)\) equal to zero and solving for \(x\). So, \(0 = (x+5)^{2} - 6\), leading to \((x+5)^{2} = 6\). Taking the square root of both sides gives \(x+5 = \pm\sqrt{6}\). Hence, the solutions are \(x = \sqrt{6} - 5\) and \(x = -\sqrt{6} - 5\), which are the x-intercepts. The y-intercept is found by setting \(x = 0\) in the function, giving \(f(0) = (0+5)^{2} - 6 = 19\). Therefore, the y-intercept is at (0,19).

Sketch the function graph

Use the identified features to sketch the graph. Start by marking the vertex at \(-5, -6\), the y-intercept at (0,19) and the x-intercepts at \(\sqrt{6} - 5\) and \(-\sqrt{6} - 5\). Since the coefficient of \((x+5)^{2}\) is positive, the graph opens upwards. Connect the points with a smooth parabolic curve.

Key concepts

Vertex Form of a Quadratic

Understanding the vertex form of a quadratic is crucial for graphing and analyzing these equations swiftly. The vertex form is expressed as \( f(x) = a(x-h)^{2} + k \), where \((h, k)\) represents the vertex of the parabola, and \(a\) determines the direction and width of the parabola.

For our exercise, the quadratic function \( f(x) = (x+5)^{2} - 6 \) is already given in vertex form. Here, \(h\) is \(-5\), and \(k\) is \(-6\), indicating the vertex of the parabola is at the point \(-5, -6\). The value of \(a\) is positive, meaning the parabola opens upward. Knowing the vertex allows us to plot the initial point and determine the shape and direction of the parabola.

X-Intercepts of a Parabola

The x-intercepts of a parabola, also known as the roots or zeros, are the points where the graph crosses the x-axis. To find these, we solve the quadratic equation \( f(x) = 0 \).

In our exercise, we set \((x+5)^{2} - 6\) equal to zero and solve for \(x\). After some algebraic manipulations, we arrive at two solutions: \(x = \(\sqrt{6} - 5\)\) and \(x = -\(\sqrt{6} - 5\)\). These solutions represent our x-intercepts, and they're essential for sketching the accurate path of the parabola across the x-axis.

Y-Intercept of a Quadratic

The y-intercept is the point where the graph intersects the y-axis. To find it, we substitute \(x = 0\) into the quadratic equation. This single intersection reveals where the graph starts or passes through the y-axis.

In the given quadratic function \((x+5)^{2} - 6\), by plugging \(x = 0\) we get \(f(0) = 25 - 6 = 19\). Therefore, the y-intercept is \((0, 19)\). It's an essential characteristic that helps in plotting the graph correctly and confirming the symmetry of the parabola.

Sketching Parabolas

Successfully sketching a parabola involves identifying key features such as the vertex, axis of symmetry, intercepts, and its direction of opening. Ensure that your sketch represents a mirror-like symmetry about the vertical axis through the vertex of the parabola.

In our example, we begin by plotting the vertex at \(-5, -6\) which gives us the axis of symmetry, which is the line \(x = -5\). Next, we plot the y-intercept at \((0, 19)\) and the x-intercepts at \(x = \(\sqrt{6} - 5\)\) and \(x = -\(\sqrt{6} - 5\)\). With these points, we can draw a smooth, U-shaped curve to represent our parabola accurately. The graph opens upwards since the coefficient of \((x+5)^{2}\) is positive. The plotted points guide the curvature, ensuring an accurate graph.