Problem 57 - 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 \((8,3) ;\) through \((10,5)\)

Step-by-Step Solution

Answer

The equation is \( P(x) = \frac{1}{2}x^2 - 8x + 35 \).

Step 1: Identify the Vertex Form

The quadratic function can be written in the vertex form: \( P(x) = a(x-h)^2 + k \). Given the vertex \((h, k) = (8, 3)\), we can substitute these into the vertex form equation: \( P(x) = a(x-8)^2 + 3 \).

Step 2: Substitute the Point Through Which the Curve Passes

The parabola passes through the point \((10, 5)\). Substitute \(x = 10\) and \(P(x) = 5\) into the vertex form equation: \( 5 = a(10-8)^2 + 3 \).

Step 3: Solve for 'a'

Simplify the equation from Step 2: \( 5 = a(2)^2 + 3 \) which simplifies to \( 5 = 4a + 3 \). Isolate \(a\) by subtracting 3 from both sides to get \(2 = 4a\). Divide by 4 to solve for \(a\): \( a = \frac{1}{2} \).

Step 4: Write the Equation in Vertex Form

Now that we have \(a = \frac{1}{2}\), substitute back into the vertex form equation: \( P(x) = \frac{1}{2}(x-8)^2 + 3 \).

Step 5: Convert Vertex Form to Standard Form

Expand the vertex form equation: \( P(x) = \frac{1}{2}(x-8)^2 + 3 \). \((x-8)^2\) expands to \(x^2 - 16x + 64\). Multiply through by \(\frac{1}{2}\): \( P(x) = \frac{1}{2}(x^2 - 16x + 64) + 3 \), simplifying to \(P(x) = \frac{1}{2}x^2 - 8x + 32 + 3\). Simplify further: \(P(x) = \frac{1}{2}x^2 - 8x + 35\).

Step 6: Verify the Derived Equation

Double-check the expanded form through calculations: The final equation derived is \( P(x) = \frac{1}{2}x^2 - 8x + 35 \). Verify that it passes through \((10, 5)\) by calculating as needed.

Key Concepts

Vertex Form

The vertex form of a quadratic function is a convenient way to express a quadratic equation when you know the vertex of the parabola it represents. The vertex form is given by the formula: \[ P(x) = a(x-h)^2 + k \]Here,

• \(a\) affects the width and direction of the parabola. A positive \(a\) opens upwards, while a negative \(a\) opens downwards.
• \((h, k)\) are the coordinates of the vertex, the highest or lowest point of the parabola.

In this form, you can easily identify the vertex, which is especially useful when graphing or when solving problems involving maxima or minima. The shift from the standard form depends on finding the right \((h, k)\) and \(a\) based on given conditions, such as passing through certain points or having a specific vertex.

Standard Form

The standard form of a quadratic function is another way to present the quadratic equation, which is often used when solving for roots or intercepts. It is expressed as:\[ P(x) = ax^2 + bx + c \]In this form:

• \(a\) is the coefficient that determines the parabola's opening direction and width. Just like in the vertex form, a positive \(a\) will open upwards, and a negative \(a\) will open downwards.
• \(b\) affects the position of the vertex along the x-axis and how steep or flat the parabola is.
• \(c\) represents the y-intercept, where the parabola crosses the y-axis.

Transitioning from vertex form to standard form involves expanding the squared term, distributing the \(a\), and simplifying the expression. This conversion is useful, as it allows direct calculation of key characteristics such as the parabola's intercepts.

Solving Quadratic Equations

Solving quadratic equations can be approached using various methods depending on the form of the given equation. Here are the common strategies:- **Factoring**: This is efficient when the quadratic can be easily expressed as a product of binomials.- **Completing the square**: This method is instrumental when the vertex form is not directly obvious but can be forced by manipulating the equation.- **Quadratic formula**: This universal method utilizes the formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) and works for any standard form equation.When you have the quadratic in vertex form, solving typically involves isolating the (\(x-h\)) term, taking the square root, and solving for \(x\). Interestingly, if you're in the standard form, transitioning back and forth can provide insights or simplify complex problems. Remember, no matter the approach, checking the solution in the context of the problem is essential to ensure accuracy.