Problem 47 - Precalculus : Building Concepts and Connections

Question

Find the vertex and axis of symmetry of the associated parabola for each quadratic function.Then find at least two additional points on the parabola and sketch the parabola by hand. $$f(t)=-16 t^{2}+100$$

Step-by-Step Solution

Answer

The vertex of the parabola is at the point (0,100), the axis of symmetry is the line \(t=0\), and two additional points on the parabola are (-1,84) and (1,84). The parabola opens downward.

Step 1: Find the vertex of the parabola

Find the t-coordinate of the vertex using the formula \(t = -b/(2a)\). There's no 'b' in this function, so it's as if \(b = 0\). Hence, the t-coordinate of the vertex is \(t=0\). Substitute \(t=0\) in the function to find the y-coordinate of the vertex, which gives \(f(0) = -16(0)^2+100 = 100\). Hence, the vertex of the parabola is at the point (0,100).

Step 2: Find the axis of symmetry of the parabola

The axis of symmetry is the vertical line passing through the vertex. The t-coordinate of the vertex is 0, so the equation of the axis of symmetry is \(t=0\).

Step 3: Find at least two additional points on the parabola

Choose two values for 't', one negative and one positive. For example, let's take \(t = -1\) and \(t = 1\). Plug these values into the function to find the corresponding 'f(t)' values. For \(t = -1\), \(f(t) = -16(-1)^2 + 100 = 84\), so the point is (-1, 84). For \(t = 1\), \(f(t) = -16(1)^2 + 100 = 84\), so the point is (1,84).

Step 4: Sketch the parabola

Plot the vertex (0,100), the points (-1, 84) and (1, 84), and the axis of symmetry \(t=0\) on a graph. Connect the points with a smooth curve to sketch the parabola. As 'a' is negative, the parabola opens downward.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point on its graph. It represents the peak or lowest point, depending on whether the parabola opens upwards or downwards. For a quadratic function of the form \( f(t) = at^2 + bt + c \), you can find the t-coordinate of the vertex using the formula \( t = -b/(2a) \). In this case, the function \( f(t) = -16t^2 + 100 \) does not have a 'b' term, meaning within this context \( b = 0 \).

To find the vertex:

• Calculate \( t = 0/(-32) = 0 \)
• Substitute \( t = 0 \) back into the function to find \( f(t) \). This gives: \( f(0) = -16(0)^2 + 100 = 100 \)

Thus, the vertex of this parabola is located at the point \((0, 100)\). Since the coefficient of \( t^2 \) is negative, the vertex is the highest point, making the parabola open downwards.

Axis of Symmetry

The axis of symmetry of a parabola is an imaginary vertical line that runs through the vertex and divides the parabola into two mirror-image halves. For any quadratic function, once the vertex's \( t \) value is known, this t-value represents the axis of symmetry.

In our function \( f(t) = -16t^2 + 100 \):

• We determined that the \( t \)-coordinate of the vertex is \( 0 \).
• Therefore, the axis of symmetry is the line \( t = 0 \).

This line can be visualized on a graph as a vertical line passing through the vertex, ensuring the parabola's shape mirrors on either side beneath this line.

Graphing Parabolas

Graphing a parabola involves plotting several key points: the vertex, additional points on either side of the vertex, and then drawing the curve through these points. Start by plotting the vertex, as it lays the foundation for the symmetry and overall flow of the parabola on the graph.

Considering \( f(t) = -16t^2 + 100 \):

• Plot the vertex at \((0, 100)\).
• Choose a point to the left of the vertex, say \( t = -1 \). Substitute into the function: \( f(-1) = -16(-1)^2 + 100 = 84 \), plot \((-1, 84)\).
• Similarly, choose a corresponding point on the right, such as \( t = 1 \): \( f(1) = 84 \), and plot \((1, 84)\).

Use these points to draw a smooth curve, ensuring it reflects the symmetry around the axis of symmetry. Since the coefficient of \( t^2 \) is negative, the parabola opens downwards, which means the curve will descend away from the vertex.