Problem 14 - Algebra 2

Question

Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b. Make a table of values that includes the vertex. c. Use this information to graph the function. $$ f(x)=x^{2}+4 $$

Step-by-Step Solution

Answer

The \(y\)-intercept is 4, axis of symmetry is \(x = 0\), vertex is (0, 4). Use these points for the graph.

Step 1: Identify the y-intercept

For the quadratic function \(f(x) = x^2 + 4\), the \(y\)-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the equation to find \(f(0)\):\[f(0) = 0^2 + 4 = 4\]Hence, the \(y\)-intercept is \(4\).

Step 2: Find the axis of symmetry and x-coordinate of the vertex

The equation of the axis of symmetry for a quadratic equation \(f(x) = ax^2 + bx + c\) is \(x = -\frac{b}{2a}\). For the function \(f(x) = x^2 + 4\), \(a = 1\) and \(b = 0\).\[x = -\frac{0}{2(1)} = 0\]Thus, the axis of symmetry and the \(x\)-coordinate of the vertex is \(x = 0\).

Step 3: Determine the vertex of the parabola

Since the \(x\)-coordinate of the vertex is 0, substitute \(x = 0\) into the function to find the \(y\)-coordinate:\[f(0) = 0^2 + 4 = 4\]Therefore, the vertex is \((0, 4)\).

Step 4: Create a table of values including the vertex

Choose a set of \(x\)-values around the vertex to complete a table of values. Include the vertex \((0, 4)\):- \(x = -2\), \(f(-2) = (-2)^2 + 4 = 8\)- \(x = -1\), \(f(-1) = (-1)^2 + 4 = 5\)- \(x = 0\), \(f(0) = 0^2 + 4 = 4\)- \(x = 1\), \(f(1) = (1)^2 + 4 = 5\)- \(x = 2\), \(f(2) = (2)^2 + 4 = 8\)The table is:\[\begin{array}{|c|c|}\hlinex & f(x) \ \hline-2 & 8 \-1 & 5 \0 & 4 \1 & 5 \2 & 8 \\hline\end{array}\]

Step 5: Graph the quadratic function

Using the table of values and the vertex, sketch the graph of the function.1. Plot the points \((-2, 8), (-1, 5), (0, 4), (1, 5), (2, 8)\).2. Draw a smooth curve through these points to show the U-shaped parabola.3. Ensure the vertex \((0, 4)\) is correctly shown as the lowest point of the parabola, which opens upwards.

Key Concepts

Understanding the y-intercept

In any quadratic function, identifying the **y-intercept** is vital as it shows where the graph meets the y-axis. In simpler terms, it's the point where the value of x is zero in the equation. To find the y-intercept for a quadratic equation like \(f(x) = x^2 + 4\), you substitute \(x = 0\) into the function. This gives you:

• \(f(0) = 0^2 + 4 = 4\)

Hence, the y-intercept for this particular function is \(4\). What this actually means is that the parabola crosses the y-axis at the point (0, 4). This is a significant point when you're plotting the graph because it's one of the few exact coordinates you know without additional calculation. Understanding this helps in providing a clear starting point for sketching the quadratic function.

What is the axis of symmetry?

The **axis of symmetry** in a quadratic function is a vertical line that splits the parabola into two identical halves. It always passes through the vertex of the parabola, acting like a mirror. Mathematically, the equation for the axis of symmetry is found using the formula \(x = -\frac{b}{2a}\) when the quadratic formula is \(ax^2 + bx + c\). For our example \(f(x) = x^2 + 4\), where \(a = 1\) and \(b = 0\), the calculation is:

• \(x = -\frac{0}{2(1)} = 0\)

This shows that the axis of symmetry for this function is the vertical line passing through \(x = 0\). This line is crucial since it not only divides the parabola into equal halves but also helps you locate the vertex, making graph sketching more precise.

Discovering the vertex of a parabola

The **vertex of a parabola** is a key feature in the graph of a quadratic function, found precisely on the axis of symmetry. It is either the highest or lowest point depending on whether the parabola opens upwards or downwards. For our quadratic function \(f(x) = x^2 + 4\), we already know the x-coordinate of the vertex is \(x = 0\) since it lies on the axis of symmetry.To find the y-coordinate, substitute \(x = 0\) back into the equation:

• \(f(0) = 0^2 + 4 = 4\)

Thus, the vertex of the parabola is located at \((0, 4)\). In this case, the graph opens upwards, making the vertex a minimum point. Knowing the vertex aids in sketching the graph accurately, as you ensure this point is represented as the lowest part of the upward-opening parabola.