Problem 45 - Calculus Single Variable

Question

Two functions \(f\) and \(g\) are given. Find constants \(h\) and \(k\) such that \(g(x)=f(x+h)+k .\) Describe the relationship between the plots of \(f\) and \(g\). \(f(x)=3 x^{2}, g(x)=3 x^{2}+12 x\)

Step-by-Step Solution

Answer

The graph of \( g(x) \) is \( f(x) \) shifted 2 units left and 12 units down.

Step 1: Understand the Relationship between f(x) and g(x)

We know that \( g(x) = f(x + h) + k \). The goal is to find the constants \( h \) and \( k \) such that this equation holds true with the given functions \( f(x) = 3x^2 \) and \( g(x) = 3x^2 + 12x \).

Step 2: Expand and Simplify f(x+h)

Substitute \( x + h \) into \( f(x) \):\[ f(x+h) = 3(x+h)^2. \]Expand the equation:\[ f(x+h) = 3(x^2 + 2hx + h^2) = 3x^2 + 6hx + 3h^2. \]

Step 3: Compare Coefficients to Solve for h

Now, compare the expanded expression for \( f(x+h) \) with \( g(x) \):\[ g(x) = f(x+h) + k = 3x^2 + 6hx + 3h^2 + k. \]Given \( g(x) = 3x^2 + 12x \), match coefficients of \( x \):Coefficient of \( x \): \( 6h = 12 \) Solve for \( h \):\[ h = \frac{12}{6} = 2. \]

Step 4: Solve for k

Now that \( h = 2 \), substitute \( h \) back into the expression:\[ g(x) = 3x^2 + 6(2)x + 3 \cdot 2^2 + k = 3x^2 + 12x + 3\cdot4 + k. \] Since \( g(x) = 3x^2 + 12x \), we have:\[ 3x^2 + 12x + 12 + k = 3x^2 + 12x. \]Match constant terms to find \( k \):\[ 12 + k = 0 \ k = -12. \]

Step 5: Conclusion and Graph Relationship

The constants are \( h = 2 \) and \( k = -12 \). Thus, \( g(x) = f(x+2) - 12 \).The graph of \( g(x) \) is a horizontal shift 2 units to the left and a vertical shift downward of 12 units compared to the graph of \( f(x) \).

Key Concepts

Quadratic Functions

Quadratic functions are fundamental in algebra and can be recognized by their standard form, which is expressed as \( f(x) = ax^2 + bx + c \). This function type is characterized by a parabolic shape when graphed on a coordinate plane. The "a" coefficient determines the "width" and orientation of the parabola, whether it opens upwards or downwards. For the function \( f(x) = 3x^2 \), the parabola opens upwards because the coefficient "a" is positive. The greater the value of "a," the "narrower" the parabola appears. Quadratics are critical as they nicely model various real-life phenomena like projectile motion. Understanding them forms a bridge to learning more complex mathematical concepts and functions.

Horizontal Shifts

Horizontal shifts involve moving a function left or right across the coordinate plane. Consider the formula \( g(x) = f(x + h) \), where "h" represents the horizontal shift. If "h" is positive, this corresponds to shifting the function left by "h" units; if "h" is negative, it shifts the function right by "|h|" units. In the context of the exercise, when we found \( h = 2 \), this indicated that the function \( f(x) = 3x^2 \) shifted 2 units to the left to become \( g(x) = 3(x+2)^2 -12 \). This horizontal movement affects only the "x" value within the function, preserving the shape but relocating the entire graph on the plane.

Vertical Shifts

Vertical shifts modify the height of a graph without altering its shape. This change is represented by the addition or subtraction of a constant "k" to the function: \( g(x) = f(x) + k \). If "k" is positive, the graph shifts upward by "k" units; if "k" is negative, it shifts downward by "|k|" units. In our example, we found \( k = -12 \), meaning that the graph of \( f(x) = 3x^2 \) moves 12 units downward to form \( g(x) = 3x^2 - 12 \). Combining vertical shifts with horizontal ones can significantly change the position of a graph while maintaining its original parabolic form. This concept is widely applied in various fields to model phenomena that experience constant shifts in baseline values.