Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-12: Express \(f(x)\) in the form \(a(x-h)^{2}+k\). $$ f(x)=-\frac{3}{4} x^{2}+9 x-34 $$

Step-by-Step Solution

Answer

The expression is \(-\frac{3}{4}(x - 6)^2 - 7\).

Step 1: Factor out the coefficient of x²

We need to rewrite the quadratic expression in a form that allows us to complete the square. Factor out the coefficient of the \( x^2 \) term, which is \(-\frac{3}{4}\). This step results in:\[f(x) = -\frac{3}{4}(x^2 -12x) - 34\]

Step 2: Complete the square

To complete the square inside the parentheses, take the coefficient of \(x\) (which is \(-12\)), halve it to get \(-6\), and then square it to get \(36\). Add and subtract \(36\) inside the parentheses:\[f(x) = -\frac{3}{4}(x^2 -12x + 36 - 36) - 34\]

Step 3: Simplify inside the parentheses

Recognize that within the parentheses we have a perfect square trinomial:\[f(x) = -\frac{3}{4}((x - 6)^2 - 36) - 34\]

Step 4: Remove the extra constant term

Distribute \(-\frac{3}{4}\) to the terms inside the parentheses:\[f(x) = -\frac{3}{4}(x - 6)^2 + \frac{3}{4} \times 36 - 34\]Calculate the constant: \[\frac{3}{4}\times36 = 27\]

Step 5: Finalize the expression

Subtract \(27\) from \(34\) to determine the value of \(k\):\[f(x) = -\frac{3}{4}(x - 6)^2 + 27 - 34\]Simplifies to:\[f(x) = -\frac{3}{4}(x - 6)^2 - 7\]

Key Concepts

Quadratic functions

Quadratic functions play a significant role in mathematics, especially in algebra. The general form of a quadratic function is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratics are pivotal because they graph as parabolas, which have a distinct U-shape.

Key features of a quadratic function include:

• The vertex, which is the highest or lowest point of the parabola depending on whether it opens upwards or downwards.
• The axis of symmetry, a vertical line that passes through the vertex, dividing the parabola into two symmetrical halves.
• Roots or x-intercepts, which are the values of \( x \) for which the function equals zero.

Understanding quadratic functions is crucial as they model many real-world phenomena, such as projectile motion or areas in geometry.

Vertex form

The vertex form of a quadratic function is a way of expressing the equation that highlights the vertex's position. It is given by \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.

This form makes it easy to identify the vertex and therefore the parabola's maximum or minimum value, depending on the orientation. Here's why vertex form is beneficial:

• It allows for instant identification of the vertex, \( (h, k) \).
• It simplifies the process of graphing the function since you immediately know the vertex location and how the parabola opens.

To convert a standard quadratic form to vertex form, completing the square is a useful technique. This involves organizing the equation to have a perfect square trinomial, making it easier to rewrite in vertex form.

Factoring quadratics

Factoring quadratics involves rewriting the quadratic expression as a product of two binomials. This method is crucial for solving quadratic equations and finding the roots. A quadratic in the form \( ax^2 + bx + c \) can sometimes be factored into \( (px + q)(rx + s) \), where \( p \), \( q \), \( r \), and \( s \) are constants.

Factoring is especially useful when dealing with equations, as it makes solving for \( x \) straightforward. Here are some key benefits of factoring quadratics:

• Efficiently finds roots or solutions of the quadratic equation, \( ax^2 + bx + c = 0 \).
• Useful for simplifying expressions and solving algebraic problems.

When a quadratic does not factor easily, completing the square or using the quadratic formula are alternative methods. In more complex cases where factoring is not possible, the quadratic formula can provide the solution.