Problem 24 - Algebra A Combined Function

Question

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=x^{2}-6 x+11 $$

Step-by-Step Solution

Answer

The vertex is (3, 2), the graph opens upward, the y-intercept is 11, and there are no real x-intercepts.

Step 1: Identify the Coefficients

For the quadratic function \(f(x) = ax^2 + bx + c\), identify \(a = 1\), \(b = -6\), and \(c = 11\). These will be used to find the vertex and determine the direction of the parabola.

Step 2: Determine the Vertex Using the Vertex Formula

The vertex \((h, k)\) of the quadratic function can be found using the formulas: \(h = -\frac{b}{2a}\) and \(k = f(h)\). Calculate \(h = -\frac{-6}{2 \cdot 1} = 3\). Substitute \(h = 3\) back into the function to find \(k\): \(f(3) = 3^2 - 6 \times 3 + 11 = 9 - 18 + 11 = 2\). So, the vertex is \((3, 2)\).

Step 3: Determine the Direction of the Parabola

Since \(a = 1\), which is positive, the parabola opens upward.

Step 4: Calculate the Y-Intercept

The y-intercept is found by evaluating the function at \(x = 0\). Plug \(x = 0\) into the function: \(f(0) = 0^2 - 6 \times 0 + 11 = 11\). So, the y-intercept is \(11\).

Step 5: Find the X-Intercept

The x-intercepts occur where \(f(x) = 0\). Solve the quadratic equation \(x^2 - 6x + 11 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculate the discriminant \(b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 11 = 36 - 44 = -8\). Since the discriminant is negative, there are no real x-intercepts.

Step 6: Sketch the Graph

Plot the vertex \((3, 2)\) and the y-intercept \((0, 11)\) on a graph. As the parabola opens upwards and does not cross the x-axis, sketch a U-shaped curve through these points that extends infinitely upwards.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a critical point, especially for graphing quadratic functions. In a standard quadratic form, given as \(f(x) = ax^2 + bx + c\), the vertex \((h, k)\) serves as either the highest or lowest point on the graph, depending on the direction the parabola opens.

• To find the vertex, we use the formula \(h = -\frac{b}{2a}\). This gives the x-coordinate of the vertex.
• Next, to find the y-coordinate \(k\), substitute \(h\) back into the original function: \(f(h) = a(h)^2 + bh + c\).

For the given function \(f(x)=x^{2}-6x+11\), first, calculate \(h = -\frac{-6}{2 \times 1} = 3\). Then, substituting \(3\) for \(x\) in the function gives \(k = 3^2 - 6 \times 3 + 11 = 2\). Thus, the vertex is at the point \((3, 2)\). This point is significant as it represents the turning point of the parabola.

Parabola Opening Direction

The direction in which a parabola opens is determined by the coefficient \(a\) in the quadratic expression \(f(x) = ax^2 + bx + c\). Here is a guideline for determining the opening direction:

• If \(a > 0\), the parabola opens upward, resembling a U-shape.
• If \(a < 0\), the parabola opens downward, resembling an inverted U, or an n-shape.

In the function \(f(x) = x^2 - 6x + 11\), the coefficient \(a = 1\), which is positive. Therefore, this parabola opens upward. Knowing the opening direction helps visualize and graph the function correctly.

Intercepts of a Quadratic

In graphing quadratics, identifying intercepts is crucial as these are the points where the graph intersects the axes.**Y-Intercept**:

• The y-intercept occurs where \(x = 0\). It is straightforward to find by evaluating \(f(0)\).

For \(f(x)=x^{2}-6x+11\), when \(x = 0\), \(f(0) = 11\). Thus, the y-intercept is the point \((0, 11)\).**X-Intercept(s)**:

• X-intercepts occur where \(f(x) = 0\). Solving the equation helps find these points.
• If the discriminant \(b^2 - 4ac\) is positive, there are two real solutions (two intercepts). If zero, one real solution (one intercept). If negative, no real solutions, thus no x-intercepts.

In our example, \(x^2 - 6x + 11 = 0\) has a discriminant of \(-8\), indicating no real x-intercepts.

Graphing Quadratics

Graphing a quadratic function gives a visual representation of the equation and shows all critical points like the vertex and intercepts.Here’s how to effectively graph a quadratic:

• Start by plotting the vertex. For \(f(x)=x^{2}-6x+11\), this is \((3, 2)\).
• Next, plot the y-intercept found at \((0, 11)\).
• Since the parabola opens upward and there are no real x-intercepts, the graph will curve back up from the vertex.

With these points, you can draw a smooth, symmetrical U-shaped curve opening upwards. The graph continues indefinitely upwards but does not cross the x-axis, consistent with the negative discriminant result.