Question

a. If \(f(x)=x^{2}-2 x+k\) and \(f(1)=3\), find \(k\). b. If \(g(t)=|t-1|+k\) and \(g(-1)=0\), find \(k\).

Short answer

\(k_{f(x)} = 4\), \(k_{g(t)} = -2\)

Step-by-step solution

Use the given information in the equation.

Since we are given that f(1) = 3, we can replace x with 1 in the equation: f(1) = (1)^2 - 2(1) + k

Solve the equation for k.

Now we can solve the equation to find the value of k: 3 = 1 - 2 + k 3 = -1 + k k = 3 + 1 k = 4 So the value of k for the function f(x) = x^2 - 2x + k is 4. b. Find k for the function g(t) = |t - 1| + k

Use the given information in the equation.

Since we are given that g(-1) = 0, we can replace t with -1 in the equation: g(-1) = |-1 - 1| + k

Solve the absolute value expression.

First, we need to solve the absolute value of the expression inside: |-1 - 1| = |-2| = 2 So the equation becomes: g(-1) = 2 + k

Solve the equation for k.

Now we can solve the equation to find the value of k: 0 = 2 + k k = -2 So the value of k for the function g(t) = |t - 1| + k is -2.

Key concepts

Absolute Value Equations

Absolute value equations involve variables within absolute value bars, indicated by \( | ... | \) in mathematical notation. The absolute value of a number is its distance from 0 on the number line, regardless of direction. Hence, \( |x| \) is always non-negative.

When solving equations like \( |t - 1| = y \), there are two cases to consider; one where \( t - 1 \) is positive (or zero) and one where \( t - 1 \) is negative. If it's positive, the equation is simply \( t - 1 = y \). If negative, you must account for the fact that \( |t - 1| \) equals \( -(t - 1) \), giving us the equation \( -(t - 1) = y \).

For the provided exercise involving \( g(t) = |t - 1| + k \) and \( g(-1) = 0 \), we first determine the absolute value \( |-1 - 1| \) which simplifies to \( |-2| = 2 \). The equation \( g(-1) = 2 + k \) follows, leading us to find that \( k = -2 \) after solving for \( k \) by subtracting 2 from both sides.

Quadratic Functions

Quadratic functions are polynomial functions with the highest power of the variable being 2, typically written as \( ax^2 + bx + c \). The graph of a quadratic function is a parabola that either opens upwards or downwards depending on the sign of the coefficient \( a \).

When solving for constants in a quadratic function like the given exercise \( f(x) = x^2 - 2x + k \) where \( f(1) = 3 \), we substitute \( x \) with 1 and set the function equal to 3. Following that, we solve for \( k \) by simplifying the equation \( 1 - 2 + k = 3 \) to find \( k = 4 \) by isolating \( k \) on one side.

Understanding the role of \( k \) in a quadratic equation is vital. It represents the y-intercept in graph form, meaning it's the point where the parabola crosses the y-axis. Hence, by determining the value of \( k \) using given points, we can fully specify the parabolic curve's position on the coordinate plane.

Function Evaluation

Function evaluation involves substituting a number for the variable in the function's expression and completing the arithmetic to arrive at a single number. It's like plugging in a specific input value and calculating the output.

When a function, for instance \( f(x) \), is given, and we need to find \( f(1) \), we replace every instance of \( x \) in the function's formula with 1. This task requires careful arithmetic work, ensuring that all operations (exponents, multiplication, addition, subtraction) are applied correctly.

In the given exercise, when we evaluate \( f(1) \) for the quadratic function \( x^2 - 2x + k \), we follow the order of operations to simplify the expression accurately. After calculation, we compare the result to the provided value of the function at \( f(1) = 3 \) and proceed to solve for the constant \( k \).

Evaluating functions is a fundamental skill in algebra that students must master, as it not only reinforces the concept of functions as input-output relationships but also sets the stage for more complex applications in calculus and other higher mathematics disciplines.