Problem 18 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-22: (a) Use the quadratic formula to find the zeros of \(f\). (b) Find the maximum or minimum value of \(f(x)\). (c) Sketch the graph of \(f\). $$ f(x)=-4 x^{2}+4 x-1 $$

Step-by-Step Solution

Answer

The zero of \(f(x)\) is \(x = \frac{1}{2}\), the maximum value is 0, and the graph is a downward parabola.

Step 1: Identifying the coefficients

We start by identifying the coefficients in the quadratic equation, which are crucial for applying the quadratic formula. The standard form of a quadratic equation is given by: \(ax^2 + bx + c = 0\). In this problem, \(a = -4\), \(b = 4\), and \(c = -1\).

Step 2: Applying the quadratic formula

The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substituting the coefficients \(a = -4\), \(b = 4\), \(c = -1\), we calculate the discriminant first: \(b^2 - 4ac = 16 - 16 = 0\). Since the discriminant is zero, there will be one real root. Applying the formula, we find \(x = \frac{-4 \pm 0}{-8} = \frac{1}{2}\).

Step 3: Finding the vertex

The parabola represented by \(f(x) = -4x^2 + 4x - 1\) has its vertex at the point \((x, f(x))\), where \(x = -\frac{b}{2a}\). Substituting \(b = 4\) and \(a = -4\), we find \(x = \frac{1}{2}\). Substituting \(x = \frac{1}{2}\) back into the equation for \(f(x)\), we calculate \(f\left(\frac{1}{2}\right) = -4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - 1 = 0\), so the vertex is \(\left(\frac{1}{2}, 0\right)\). This is also the maximum point because the parabola opens downwards (\(a < 0\)).

Step 4: Sketching the graph

To sketch the graph of \(f(x) = -4x^2 + 4x - 1\), plot the vertex \(\left(\frac{1}{2}, 0\right)\) and note that it is a parabola opening downwards, given that \(a = -4\) is negative. The vertex is also where it touches the x-axis because the zero is found at \(x = \frac{1}{2}\). This implies the parabola does not cross the x-axis, as it only touches at the vertex.

Key Concepts

Quadratic Formula

When faced with a quadratic equation like \( f(x) = -4x^2 + 4x - 1 \), calculating the zeros can be straightforward using the quadratic formula. This formula roots from the general form \( ax^2 + bx + c = 0 \). It provides a powerful method for finding solutions to quadratic equations: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• Where \( b^2 - 4ac \) is called the discriminant.
• The discriminant determines the nature of the roots.

Knowing your coefficients and substituting them into this formula, you can discover the values for \( x \) where the quadratic function crosses the x-axis (the "zeros" or "roots"). In our example equation, after inserting the values, we find one real root at \( x = \frac{1}{2} \), indicating our parabola just touches the x-axis without crossing it, due to the discriminant being zero. This indicates a single, repeated root.

Vertex of a Parabola

Finding the vertex of a parabola, especially one defined by \( f(x) = -4x^2 + 4x - 1 \), reveals the highest or lowest point on the graph of the quadratic equation. The vertex is important because it provides valuable information about the graph's shape and orientation.

• The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \).
• For the given equation, substituting \( b = 4 \) and \( a = -4 \) results in \( x = \frac{1}{2} \).

Calculate the function's value at this x-coordinate to find the full vertex: \( f\left(\frac{1}{2}\right) = 0 \), giving the vertex point as \( \left(\frac{1}{2}, 0\right) \). Since the quadratic equation has \( a = -4 \), which is negative, the parabola opens downwards, establishing the vertex as a maximum point on the graph. This point determines the extremity of the parabola's arch.

Parabola Graphing

Plotting the graph of quadratic functions requires identifying key features, such as the vertex and axis of symmetry. The equation \( f(x) = -4x^2 + 4x - 1 \) defines a parabola, a U-shaped curve that can open either upwards or downwards. For our function:

• The parabola opens downwards due to the negative coefficient \( a = -4 \).
• The vertex \( \left(\frac{1}{2}, 0\right) \) serves as a pivot, being the highest point.

Since the parabola doesn't have any additional intersections with the x-axis—the vertex is also where it just touches the x-axis—the graph doesn’t cross over. Understanding these components assists in accurately sketching the curve, clearly showing its orientation and vertex position. Remember, adding a dashed line along the axis of symmetry, \( x = \frac{1}{2} \), helps depict the balance that symmetrically splits the parabola.