Problem 71 - Precalculus : Building Concepts and Connections

Question

Let \(g(s)=-2 s^{2}+b s .\) Find the value of \(b\) such that the vertex of the parabola associated with this function is (1,2)

Step-by-Step Solution

Answer

The value of \(b\) such that the vertex of the parabola associated with the function \(g(s)=-2 s^2 +bs\) is (1,2) is -4.

Step 1: Formulate equation for x-coordinate of the vertex

The x-coordinate of the vertex is given by \(-b/2a\). Plugging in \(x = 1\) (the given x-coordinate of the vertex) and \(a = -2\), gives \(-b/2(-2) = 1\).

Step 2: Solve for b

Resolve this equation to find \(b\). Multiplying both sides by \(-4\) gives \(b = -4\). Thus the value of \(b\) which ensures the vertex of the parabola is at (1,2), is -4.

Step 3: Verification

Verify this result by substituting \(b = -4\) and \(s = 1\) into the equation \(g(s)= -2 s^2 + bs\). This should yield \(g(1) = 2\) to confirm that the y-coordinate of the vertex is indeed 2.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that is specifically concerned with terms up to the second degree. It usually takes the form: \[ f(x) = ax^2 + bx + c \] where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The shape of the graph of a quadratic function is a parabola. This means that it has a unique U-shaped curve.

• The value of \(a\) determines if the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)).
• The value of \(b\) affects the direction and steepness of the parabola's sides.
• The value of \(c\) represents the y-intercept, which is the point where the parabola crosses the y-axis.

Understanding these parameters is crucial for sketching the graph of a quadratic function and predicting its behavior.

Vertex Formula

The vertex of a parabola formed by a quadratic equation provides a key point of information. It serves as either the highest or lowest point on the graph, depending on whether the parabola opens downwards or upwards, respectively. The vertex formula helps us in finding the coordinates of this point.
The x-coordinate of the vertex for the standard quadratic function \(f(x) = ax^2 + bx + c\) is calculated using the formula:\[ x = \frac{-b}{2a} \]
Once the x-coordinate is determined, the corresponding y-coordinate can be found by substituting\( x \) back into the original equation. In this exercise, it's given that the vertex is \((1, 2)\). By substituting \(x = 1\) and \(a = -2\) into the vertex formula and solving for \(b\), we can identify how \(b\) affects the position of the vertex.

Solving for Coefficients

Sometimes, you might be given specific conditions, such as a specific vertex, and need to determine one of the coefficients in the quadratic function. This involves a few steps:

• Use the vertex formula \(x = \frac{-b}{2a}\) to find a relation between the coefficients using the given vertex x-coordinate.
• Substitute this x-value into the parabola equation to verify with the y-coordinate of the vertex.

In our specific exercise, after using the relation from the x-coordinate of the vertex \((x=1)\), we determined that \(b = -4\).
Verification involves plugging these values back in to ensure the y-coordinate of the vertex is correct. Hence, by inputting \(s = 1\) into the equation \(g(s) = -2s^2 + bs\) with \(b = -4\), it confirms that \(g(1) = 2\), thus proving that the parabola's vertex is indeed at \((1, 2)\).