Problem 73 - Precalculus : Building Concepts and Connections

Question

Sketch a graph of the quadratic function, indicating the vertex, the axis of symmetry, and any \(x\)-intercepts. $$F(s)=-2 s^{2}+3 s+1$$

Step-by-Step Solution

Answer

In summary, the graph of the function \(f(x) = -2x^2 + 3x +1\) is a downward-opening parabola with vertex (0.75, 1.125), axis of symmetry \(x = 0.75\), and x-intercepts \((-0.25, 0)\) and \((1, 0)\).

Step 1: Calculate the Vertex

This is the point \((-b/2a, f(-b/2a))\). In this case, \(a=-2, b=3, c=1\), therefore the vertex is at \(x=-b/2a= -3/-4 = 0.75\). When we substitute \(x=0.75\) into the equation, we find that \(f(0.75) = -2(0.75)^2 + 3(0.75) + 1 = 1.125\). So, the vertex is at (0.75, 1.125).

Step 2: Determine the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex, so in this case it is \(x=0.75\).

Step 3: Calculate the x-Intercepts

The x-intercepts are the points where the function intersects with the x-axis (where \(f(x)=0\)). To solve for x when \(f(x)=0\), we use the quadratic formula: \(x=(-b±sqrt(b^2-4ac))/(2a)\). In this case, applying the given coefficients to the quadratic formula we find that \(x=(3±sqrt(3^2-4*-2*1))/2*-2=(3±sqrt(9+8))/4=(-0.5, 1)\). So, the x-intercepts are at \((-0.25, 0)\) and \((1, 0)\).

Step 4: Sketch the Graph

Plot the vertex. Then plot the x-intercepts and the axis of symmetry. Since the function is a downward-opening parabola (since \(a<0\)), sketch the graph is done by constructing a parabola shaped curve, opening downwards and going through the plotted points.

Key Concepts

Vertex of a Parabola

In quadratic functions, the vertex is a crucial point. Think of it as the peak or the lowest point of a curve, called a parabola. For the function you're working with, the vertex can be found using the formula for the x-coordinate:

• denoted by \(-\frac{b}{2a}\). This gives us the horizontal position of the vertex.
• Then you insert this value back into the function to find the y-coordinate.

This approach helps in analyzing the function, as knowing the vertex tells us the maximum or minimum value of the quadratic function. In your exercise, the vertex for the quadratic function \(F(s)=-2s^2+3s+1\) is calculated to be at the point \((0.75, 1.125)\). This particular vertex suggests that the parabola opens downwards, with the vertex being the maximum point.

Axis of Symmetry

The **axis of symmetry** in a quadratic function is an imaginary vertical line. It runs through the vertex and divides the parabola into two mirror-image halves. In essence, it helps balance the curve to one side just as much as the other side. You can think of it as the line upon which the parabola flips perfectly.

For any quadratic function given by \( ax^2 + bx + c \), you determine the axis of symmetry using the formula:

• \(x = \frac{-b}{2a} \).

This same value was used to find the vertex's x-coordinate, because the axis of symmetry passes directly through it.
In the original exercise, \( F(s) = -2s^2 + 3s + 1 \), the line \(x=0.75\) serves as the axis of symmetry. It's a helpful reference when sketching or analyzing the graph.

Quadratic Formula

The quadratic formula is an essential tool in solving equations of the form \(ax^2 + bx + c = 0\). It allows you to identify the roots, or x-intercepts, where the parabola crosses the x-axis. The formula is expressed as:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, the symbol \( \pm \) indicates that there will usually be two solutions—showing where the parabola intersects the x-axis. If those roots are real numbers, your graph will clearly show two points where the curve crosses the axis. In complex number cases, the graph doesn't touch the x-axis.

For the step in your exercise, the task was to apply the quadratic formula to \(F(s)=-2s^2+3s+1\), which results in x-intercepts at \((-0.25, 0)\) and \((1, 0)\). Knowing these intercepts helps in sketching the curve of the parabola accurately, ensuring that it captures all critical points.