Problem 58 - A Graphical Approach to College Algebra

Question

Find the equation of the quadratic function satisfying the given conditions. (Hint: Determine values of \(a\), \(h,\) and \(k\) that satisfy \(P(x)=a(x-h)^{2}+k .\) ) Express your answer in the form \(P(x)=a x^{2}+b x+c .\) Use your calculator to support your results. Vertex \((-6,-12)\); through \((6,24)\)

Step-by-Step Solution

Answer

The equation is \(P(x) = \frac{1}{4}x^2 + 3x - 3\).

Step 1: Identify the Vertex Form

The vertex of the parabola is given as \((-6, -12)\). The vertex form of a quadratic equation is \(P(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex. Thus, substitute \(h = -6\) and \(k = -12\) into the equation, so we have \(P(x) = a(x+6)^2 - 12\).

Step 2: Plug in the Given Point

The quadratic passes through the point \((6, 24)\). Substitute \(x = 6\) and \(P(x) = 24\) into the equation from Step 1. This gives us the equation: \(24 = a(6+6)^2 - 12\).

Step 3: Solve for 'a'

Simplify the equation from Step 2: \(24 = a(12)^2 - 12\) which becomes \(24 = 144a - 12\). Add 12 to both sides to get \(36 = 144a\). Divide both sides by 144 to find \(a = \frac{1}{4}\).

Step 4: Write the Equation in Vertex Form

Plug \(a = \frac{1}{4}\) back into the vertex form equation: \(P(x) = \frac{1}{4}(x+6)^2 - 12\).

Step 5: Convert to Standard Form

Expand \((x+6)^2\) to get \(x^2 + 12x + 36\). Substitute back into the equation: \(P(x) = \frac{1}{4}(x^2 + 12x + 36) - 12\). Distribute \(\frac{1}{4}\) to get \(P(x) = \frac{1}{4}x^2 + 3x + 9 - 12\). Simplify to \(P(x) = \frac{1}{4}x^2 + 3x - 3\).

Step 6: Finalize the Equation

The equation of the quadratic function in standard form is \(P(x) = \frac{1}{4}x^2 + 3x - 3\).

Key Concepts

Vertex Form of a Quadratic Function

The vertex form of a quadratic function is a unique way to express the equation of a parabola. The form is given by \( P(x) = a(x-h)^2 + k \), where \((h, k)\) denotes the vertex of the parabola. In this form, \(h\) represents the horizontal shift from the origin, and \(k\) represents the vertical shift. This form is particularly useful when you know the vertex of the parabola and need to write the equation.

To better understand it:

• The vertex \((h, k)\) is the point where the parabola reaches its maximum or minimum value.
• The parameter \(a\) determines the direction and the width of the parabola.
• If \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards.

In the exercise, the vertex is at \((-6, -12)\), so substituting \(h = -6\) and \(k = -12\) into the vertex form leads to \( P(x) = a(x + 6)^2 - 12 \). This allows us to find other unknowns easily if another point on the parabola is provided.

Standard Form of a Quadratic Function

The standard form is another way to express a quadratic equation, written as \( P(x) = ax^2 + bx + c \). This form is valuable for analysis and solving quadratic equations, making it ideal for finding zeroes or intercepts of the function.

Features of the standard form include:

• The coefficient \(a\) indicates the parabola's direction and width.
• The coefficients \(b\) and \(c\) influence the parabola's position along the x-axis and its y-intercept, respectively.
• This form can be directly used for algebraic operations like addition, subtraction, or factoring.

In the exercise, once we found \(a = \frac{1}{4}\) from the vertex form, we plugged it into the vertex equation to transform \( (x+6)^2 \) into \( x^2 + 12x + 36 \) and distributed \( \frac{1}{4} \). The equation expanded and simplified into \( P(x) = \frac{1}{4}x^2 + 3x - 3 \). This is the standard form needed for broader applications.

Finding 'a' in Quadratics

Finding the coefficient \(a\) in quadratic equations is crucial because it dictates the parabola's basic shape and direction. To find \(a\), you need additional information besides the vertex, such as another point through which the parabola passes.

In this problem, we use point \((6,24)\) along with the vertex form \( P(x) = a(x+6)^2 - 12 \). By substituting \(x = 6\) and \(P(x) = 24\) into the equation, we can solve for \(a\):

• Replace \(P(x)\) with 24: \( 24 = a(6+6)^2 - 12 \).
• Simplify the equation to \(24 = 144a - 12\), then add 12 to both sides to get \(36 = 144a\).
• Solve for \(a\) by dividing both sides by 144, which yields \(a = \frac{1}{4}\).

The value of \(a\) is now identified, allowing us to write the quadratic equation in both vertex and standard forms, capturing the parabola's accurate trajectory.