Problem 76 - Precalculus : Building Concepts and Connections

Question

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

Step-by-Step Solution

Answer

The vertex of the function is at (0, -1), its axis of symmetry is at t=0, and there are no x-intercepts. It is a downward-opening parabola that lies above the x-axis.

Step 1: Calculate the vertex

The formula for the vertex is \((h,k)\), where \(h=-b/2a\) and \(k=f(h)\). In the function \(f(t)=-t^{2}-1\), \(a=-1\) and \(b=0\), so \(h=-0/(-2*-1)=0\), and \(k=f(0)=-0^{2}-1=-1\). Therefore, the vertex is at \((0, -1)\).

Step 2: Determine the axis of symmetry

The axis of symmetry is a vertical line passing through the vertex of the parabola which is \(t=h\), so for this function the axis of symmetry is at \(t=0\).

Step 3: Find any \(x\)-intercepts

The \(x\)-intercepts are found by setting the function equal to zero and solving for \(t\). But for the function \(f(t)=-t^{2}-1\), it is not possible to get \(f(t)=-t^{2}-1=0\), as \(t^{2}\) is always greater than or equal to zero for real numbers. Therefore, this graph does not cross the x-axis, meaning there are no \(x\)-intercepts.

Step 4: Sketch the graph

Sketch a downward-opening parabola with its maximum point at \((0, -1)\) and its axis of symmetry at \(t=0\). As there are no \(x\)-intercepts, the graph will be above the x-axis.

Key Concepts

Vertex of a Parabola

In any quadratic function, the vertex is an important point, as it can show either the highest or lowest point of the graph. For the function \( f(t)=-t^{2}-1 \), the vertex acts as the maximum point, because the parabola opens downward. The standard form of a quadratic function \( ax^2 + bx + c \) reveals the vertex via the formula \( (h,k) \). Here:

• \( h = \frac{-b}{2a} \)
• \( k = f(h) \)

In the exercise, \( a = -1 \) and \( b = 0 \), leading to \( h = 0 \) and evaluating \( k \), or \( f(0) \), gives \( -1 \). Hence, the vertex is at \( (0, -1) \). By understanding where the vertex falls on the graph, students can easily identify the key point around which the parabola is symmetrical, which aids in accurately sketching and predicting the graph's behavior.

Axis of Symmetry

The axis of symmetry for a parabola is a vertical line that passes through the vertex. It's a fundamental aspect because it divides the parabola into two mirrored halves. This line helps in understanding the parabola's symmetrical properties. Given our function, \(f(t)=-t^{2}-1\), with a vertex at \( t = 0 \), the axis of symmetry is exactly at \( t = 0 \).

• To find the axis of symmetry, rely on the formula: \( t = h \), where \( h \) is derived from the vertex formula \( h = \frac{-b}{2a} \).

For any quadratic function in which \( b = 0 \), like this one, the axis will naturally be the \( y \)-axis. This information is crucial for both manually sketching the graph and setting computational boundaries for calculus operations.

Graphing Quadratic Functions

Graphing quadratic functions is all about understanding the movements and shape of the parabola. In our case, the quadratic function \(f(t)=-t^{2}-1\) results in a parabola that opens downwards. Starting with the vertex at \( (0, -1) \), it's important to consider the overall shape and position.

• Since \(a = -1\), the parabola opens downwards, rooted at its maximum \( (0, -1) \).
• No real \(x\)-intercepts exist, as no real number squared plus one ever equals zero.

This provides a clear picture of the function above the \(x\)-axis, representing a curve pointing downward from the vertex. These insights simplify both sketching the function and predicting its trajectory. Each step concentrates on establishing the graph in relation to its axis of symmetry and vertex, ensuring a thorough understanding of its properties.