Problem 16 - Introductory Algebra for College Students

Question

Find the \(y\) -intercept for the parabola whose equation is given. $$y=x^{2}+8 x+14$$

Step-by-Step Solution

Answer

The y-intercept of the given parabola is 14.

Step 1: Identify the Quadratic Equation

The provided equation is a quadratic equation given as \( y = x^{2} + 8x + 14 \)

Step 2: Substituting x=0 in the equation

Substitute x=0 into the equation \( y = (0)^{2} + 8*(0) + 14 \)

Step 3: Calculate the value of y

On solving the equation we get \( y = 14 \).

Key Concepts

Quadratic Equation

The quadratic equation is a type of polynomial equation of the second degree, which means it includes a term with the variable raised to the power of two, commonly written in the form of

\[ ax^2 + bx + c = 0 \]
where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. In our exercise, the quadratic equation is\( y = x^2 + 8x + 14 \), where \( a = 1 \), \( b = 8 \), and \( c = 14 \).

These equations are paramount in algebra and appear in various applications across physics, engineering, and economics due to their natural occurrence in problems involving area, optimization, and motion.

Parabola

A parabola is a symmetric curve on a plane that is defined as the set of all points that are equidistant from a fixed point known as the focus and a fixed line known as the directrix. In the context of quadratic equations, the graph of a function of the form \( y = ax^2 + bx + c \) is a parabola. The orientation (upward or downward) of the parabola depends on the sign of the \( a \) term in the quadratic equation. Since our original equation has a positive \( a \) value, it means the parabola opens upwards.

Regarding the y-intercept, it is the point where the parabola crosses the y-axis, and it occurs where \( x = 0 \). This is simple to find because, at this point, both the \( x^2 \) and \( bx \) terms will equal zero, and only the constant term \( c \) will contribute to the y-value.

Algebraic Solutions

Algebraic solutions involve finding the values for variables that satisfy an equation. For linear equations, one is usually solving for a single value, whereas for quadratic equations, there can be two solutions, also known as roots. The process of finding these solutions requires algebraic manipulation.

In our textbook exercise, the y-intercept can be considered an algebraic solution where we are solving for the value of \( y \) when \( x = 0 \). This is a particular case where the algebra is straightforward because setting \( x = 0 \) simplifies the equation significantly. The algebraic method used in the solution involved directly substituting \( x \) with 0, which gave us the constant term \( c \) as the y-intercept of the parabola, demonstrating the elegant interplay between algebra and geometry in quadratic functions.