Problem 75 - Precalculus : Building Concepts and Connections

Question

Sketch a graph of the quadratic function, indicating the vertex, the axis of symmetry, and any \(x\)-intercepts. $$g(t)=t^{2}+t+1$$

Step-by-Step Solution

Answer

The vertex of the graph is at \(-\frac{1}{2}, \frac{3}{4}\), and the axis of symmetry is \(t =-\frac{1}{2}\). Due to a negative discriminant \(-3\), the graph has no x-intercepts.

Step 1: Prepare the function into vertex form

The quadratic function is currently in the standard form \(f(t)=at^{2}+bt+c\). To change it to vertex form which is \(f(t)=a(t-h)^2 +k\), where (h,k) is the vertex of the parabola, we complete the square. \n The given function is \(g(t)=t^{2}+t+1\), this can be rewritten as \(g(t)=(t^2+t)+1 = (t+\frac{1}{2})^2-\frac{1}{4}+1 = (t+\frac{1}{2})^2+\frac{3}{4}\)

Step 2: Identify the vertex and axis of symmetry

By comparing our transformed function with the vertex form, we can identify the vertex of the parabola as \(-\frac{1}{2},\frac{3}{4}\). The axis of symmetry is the vertical line containing the vertex, which here is x = -\frac{1}{2}

Step 3: Identify x-intercepts

The x-intercepts are the values of t for which g(t) = 0. Thus, we have to solve the following equation: \(0 = t^{2}+t+1\). Calculating the discriminant we have \(\Delta=b^2-4ac=1-4= -3\). Since the discriminant is less than 0, there are no real roots and thus no x-intercepts.

Key Concepts

Vertex Form of a Parabola

The vertex form of a parabola is an invaluable tool for graphing quadratic functions and understanding their properties. It is written as
\( f(x) = a(x-h)^2 + k \),
where \( (h, k) \) represents the vertex, which is the highest or lowest point of the parabola. The variable \( a \) indicates the parabola's direction and the width of its opening. If \( a \) is positive, the parabola opens upwards, and if negative, it opens downwards.

To convert a quadratic function from the standard form \( ax^2 + bx + c \) into vertex form, a method called completing the square is used.

Completing the Square

This process involves forming a perfect square trinomial from the quadratic portion of the function, which is then simplified into the vertex form.

In the given exercise, the function \( g(t)=t^{2}+t+1 \) is transformed into \( g(t)=(t+\frac{1}{2})^2+\frac{3}{4} \), revealing the vertex \( (-\frac{1}{2}, \frac{3}{4}) \). This pivotal point on the graph is a clear starting location when plotting the curve and is crucial in understanding the parabola's shape and position in the coordinate system.

Axis of Symmetry

Every parabola has a mirror-like feature, dividing it into two perfectly symmetrical halves. This feature is known as the axis of symmetry.

For the vertex form of a parabola \( f(x) = a(x-h)^2 + k \), the axis of symmetry is a vertical line that passes through the vertex at the coordinate \( x = h \). The axis of symmetry not only helps in graphing the parabola but also offers insights into the function's geometrical properties.

Finding the Axis of Symmetry

In the context of the provided exercise where the vertex is \( (-\frac{1}{2}, \frac{3}{4}) \), the axis of symmetry is the line \( x = -\frac{1}{2} \). When sketching a graph of the quadratic function, this axis serves as a guideline, ensuring that the parabola is correctly reflected on either side. It is a fundamental concept that, alongside the vertex, defines the precise shape and location of the parabola.

X-Intercepts of a Parabola

The x-intercepts of a parabola, also known as its roots or zeroes, are the points at which the graph crosses the x-axis. To find these intercepts, one must solve the equation \( f(x) = 0 \).

The quadratic formula, \( x = [-b \br\br2\frac{4ac}{2a}] \), derived from the standard form of a quadratic equation, is commonly used to calculate the x-intercepts. However, the discriminant, \( \br D = b^2 - 4ac \), plays a crucial role in determining their nature.

Understanding the Discriminant

If \( D > 0 \), there are two distinct real roots, if \( D = 0 \), there is one real root, and if \( D < 0 \), there are no real roots; the parabola does not intersect the x-axis. In our exercise, with \( g(t)=t^{2}+t+1 \), we find the discriminant to be \( -3 \), implying that there are no x-intercepts as the parabola does not cut across the x-axis. This information is essential for accurately sketching the quadratic graph and understanding its behavior.