Problem 60 - 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 \((5,6) ;\) through \((1,-6)\)

Step-by-Step Solution

Answer

The quadratic equation is \( P(x) = -\frac{3}{4}x^2 + \frac{15}{2}x - \frac{51}{4} \).

Step 1: Understand the vertex form of the quadratic equation

The vertex form of a quadratic equation is given by \( P(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex of the parabola. In this problem, the vertex is \((5, 6)\), so we substitute \(h = 5\) and \(k = 6\) into the vertex form: \[ P(x) = a(x - 5)^2 + 6 \]

Step 2: Substitute the known point to find 'a'

The quadratic passes through the point \((1, -6)\). Substitute \(x = 1\) and \(P(x) = -6\) into the equation to solve for \(a\):\[ -6 = a(1 - 5)^2 + 6 \]Simplify:\[ -6 = a(16) + 6 \]Subtract 6 from both sides:\[ -12 = 16a \]

Step 3: Solve for 'a'

Divide both sides of the equation by 16 to solve for \(a\):\[ a = -\frac{12}{16} \]Simplify the fraction:\[ a = -\frac{3}{4} \]

Step 4: Write the complete vertex form equation

Substitute \(a = -\frac{3}{4}\), \(h = 5\), and \(k = 6\) back into the vertex form:\[ P(x) = -\frac{3}{4}(x - 5)^2 + 6 \]

Step 5: Expand to get the standard form

Expand \((x - 5)^2\) to convert to the standard form:\[ (x - 5)^2 = x^2 - 10x + 25 \]Substitute this into the vertex form:\[ P(x) = -\frac{3}{4}(x^2 - 10x + 25) + 6 \]Distribute \(-\frac{3}{4}\):\[ P(x) = -\frac{3}{4}x^2 + \frac{30}{4}x - \frac{75}{4} + 6 \]Convert \(6\) to \(\frac{24}{4}\) to combine:\[ P(x) = -\frac{3}{4}x^2 + \frac{30}{4}x - \frac{75}{4} + \frac{24}{4} \]Combine the constants:\[ P(x) = -\frac{3}{4}x^2 + \frac{30}{4}x - \frac{51}{4} \]

Step 6: Express the equation in a simpler form

Simplify the expressions in terms of common fractions:\[ P(x) = -\frac{3}{4}x^2 + \frac{15}{2}x - \frac{51}{4} \]

Key Concepts

Vertex Form

The vertex form of a quadratic equation is a powerful tool for identifying the features of a parabola quickly. It is given by the formula \( P(x) = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex point of the parabola, and \( a \) determines the direction and width of the parabola.

By having the vertex in this form, we know exactly where the parabola's highest or lowest point is. In our exercise, the vertex is \((5, 6)\), meaning the highest or lowest point of the parabola is at \( x = 5 \) and \( y = 6 \). Whether it is a peak or a valley depends on the value of \( a \), with \( a < 0 \) indicating the parabola opens downwards, which is the case in our exercise.

• To find \( a \), substitute another known point of the quadratic into the vertex form.
• Remember that \( a \) influences how "steep" or "wide" the parabola looks.

This is crucial for transitioning calculations to the standard form, which further helps in understanding the properties of the quadratic function.

Standard Form

The standard form of a quadratic equation is expressed as \( P(x) = ax^2 + bx + c \). It's the form that's commonly seen and useful for analyzing the roots and intercepts of a parabola.

Converting from vertex form to standard form involves expanding and simplifying the expression. In our example, we start with \( P(x) = -\frac{3}{4}(x - 5)^2 + 6 \), which is expanded to reveal its standard components. This requires calculating \((x - 5)^2\), multiplying it by \(-\frac{3}{4}\), and then combining with constant terms.

• The process involves basic algebra: expanding brackets and combining like terms.
• Every term \( ax^2 \), \( bx \), and \( c \) plays a role in the graph's shape and orientation.

By the end, you have \( P(x) = -\frac{3}{4}x^2 + \frac{15}{2}x - \frac{51}{4} \), a neat representation that reflects both the direction and position of your parabola relative to the coordinate plane.

Solving Quadratic Equations

Solving quadratic equations often means finding the values of \( x \) (commonly known as roots) for which the quadratic \( P(x) \) equals zero. These points are crucial, representing where the parabola pierces the x-axis.

This can be executed by various methods such as factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Each method has its own advantages and can be chosen based on the specific quadratic equation in question.

• The discriminant \( b^2 - 4ac \) helps in predicting the nature of roots (real or complex). If positive, expect two distinct real solutions; if zero, one real solution; and if negative, two complex solutions.
• Using the quadratic formula is often the go-to method when factoring isn't straightforward.

Understanding the vertex and standard forms provides a fundamental basis for these calculations, enabling you to switch from examination to solution of quadratics with ease.