Problem 20 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(g(x)=\frac{1}{4} x^{2}-1\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(g(x)=\frac{1}{4}x^2-1\) is at point \((0, -1)\). Its axis of symmetry is \(x=0\). The \(x\)-intercepts are \(x=2\) and \(x=-2\), while the \(y\)-intercept is \(g(x)=-1\). To graph the function, plot these points and draw an upward-opening parabola that passes through them.

Step 1: Identify the vertex

The vertex formula for a quadratic function in the form \(g(x)=ax^2+bx+c\) is: \(h = -\frac{b}{2a}\) and \(k = g(h)\) Since in our case, \(a=\frac{1}{4}\) and \(b=0\), we have: \(h = -\frac{0}{2(\frac{1}{4})} = 0\) Then, we find k-value by plugging h back into the function: \(k = g(0) = \frac{1}{4}(0)^2 - 1 = -1\) So, the vertex of the quadratic function is at \((h,k) = (0, -1)\).

Step 2: Find the axis of symmetry

The axis of symmetry is a vertical line that passes through the vertex. Since the vertex is at \((0, -1)\), the axis of symmetry is \(x=0\).

Step 3: Calculate the x- and y-intercepts

To find the x-intercept(s), set \(g(x)=0\) and solve for \(x\): \(\frac{1}{4}x^2 - 1 = 0\) \(x^2 = 4\) \(x = \pm \sqrt{4}\) The x-intercepts are \(x=2\) and \(x=-2\). To find the y-intercept, set \(x=0\) and solve for \(g(x)\): \(g(0) = \frac{1}{4}(0)^2 - 1 = -1\) The y-intercept is \(g(x) = -1\).

Step 4: Graph the function

We can graph the function with the following points: - Vertex: \((0, -1)\) - Axis of symmetry: \(x=0\) - x-intercepts: \(x=2\) and \(x=-2\) - y-intercept: \(g(x) = -1\) When graphing, you should plot the vertex, x- and y-intercepts, and the axis of symmetry. Then draw a parabola that goes through the plotted points. The parabola should be opening upward since \(a=\frac{1}{4}\) is positive.

Key Concepts

Vertex

In a quadratic function, the vertex is a crucial point termed as the 'turning point' of the parabola. It represents the maximum or minimum value of the function. For the quadratic function given as \( g(x) = \frac{1}{4}x^2 - 1 \), the vertex provides valuable information about the graph's overall shape and position. The formula to find the vertex, specifically its horizontal component \( h \), is given by:

• \( h = -\frac{b}{2a} \)

Here, \(a = \frac{1}{4}\) and \(b = 0\), making \( h = 0 \). Once \( h \) is determined, substitute it back into the function to find \( k \):

• \( k = g(h) = \frac{1}{4}(0)^2 - 1 = -1 \)

Thus, the vertex is found at the point \((0, -1)\). This point gives a visual indication of where the parabola "turns" on the graph.

Axis of Symmetry

The axis of symmetry in a quadratic function is an imaginary vertical line that passes through the vertex. It acts like a mirror, ensuring the parabola is symmetric on either side. For the quadratic function \( g(x) = \frac{1}{4}x^2 - 1 \), the axis of symmetry can be directly derived from the vertex. Since the vertex is at \((0, -1)\), the axis of symmetry is given as:

• \( x = 0 \)

This means that the parabola is perfectly mirrored along the line \( x = 0 \). When graphing, this line helps maintain the balance of the parabola shape and ensures each side of the parabola looks identical.

x-intercepts

An important aspect of analyzing a quadratic function is locating its x-intercepts, the points where the parabola crosses the x-axis. To find these intercepts, set the function equal to zero and solve for \( x \). For the given function \( g(x) = \frac{1}{4}x^2 - 1 \), the solution is found by solving the equation:

• \( \frac{1}{4}x^2 - 1 = 0 \)
• \( x^2 = 4 \)
• \( x = \pm 2 \)

Thus, the x-intercepts are at points \( x = 2 \) and \( x = -2 \). These intercepts are essential for graphing as they show where the curve of the parabola crosses the x-axis, providing a reference for the parabola's width and position.

y-intercepts

The y-intercept is a crucial piece of information when graphing any function, as it indicates the point where the graph crosses the y-axis. For our quadratic function \( g(x) = \frac{1}{4}x^2 - 1 \), finding the y-intercept involves setting \( x = 0 \) and calculating \( g(x) \):

• \( g(0) = \frac{1}{4}(0)^2 - 1 = -1 \)

So, the y-intercept is at the point \( (0, -1) \). It coincides with the vertex of our parabola, which is not always the case but highlights the minimum point for this particular function. Knowing the y-intercept helps in sketching the graph's starting point on the vertical axis, providing further clarity on the shape of the quadratic.