Question

Consider the function \(f(x)=(x+4)(x-3)\) Without graphing, determine the zeros of the function after each transformation. a) \(y=4 f(x)\) b) \(y=f(-x)\) c) \(y=f\left(\frac{1}{2} x\right)\) d) \(y=f(2 x)\)

Short answer

a) \(-4, 3\); b) \(-4, 3\); c) \(-8, 6\); d) \(-2, \frac{3}{2}\)

Step-by-step solution

Identify the original zeros

To find the zeros of the original function, set \(f(x) = (x+4)(x-3) = 0\). Solving for \(x\), we get the zeros \(x = -4\) and \(x = 3.\)

Transformation a: Scaling the function vertically

For the transformation \(y = 4f(x)\), the function is simply scaled vertically by a factor of 4. The zeros of the function will remain the same. Hence, the zeros are still \(x = -4\) and \(x = 3.\)

Transformation b: Reflecting the function across the y-axis

For the transformation \(y = f(-x)\), replace \(x\) with \(-x\) in the function. Therefore, \(f(-x) = (-x + 4)(-x - 3)\). Solving for \(-x + 4 = 0\) and \(-x - 3 = 0\), we get the zeros \(x = -4\) and \(x = 3\).

Transformation c: Horizontal stretching

For the transformation \(y = f\left(\frac{1}{2}x\right)\), replace \(x\) with \(\frac{1}{2}x\). Therefore, \(f\left(\frac{1}{2}x\right) = \left(\frac{1}{2}x+4\right)\left(\frac{1}{2}x-3\right)\). Solving \(\frac{1}{2}x + 4 = 0\) and \(\frac{1}{2}x - 3 = 0\), multiplying both sides by 2, we get \(x = -8\) and \(x = 6\).

Transformation d: Horizontal compression

For the transformation \(y = f(2x)\), replace \(x\) with \(2x\). Therefore, \(f(2x) = \left(2x + 4\right)\left(2x - 3\right)\). Solving \(2x + 4 = 0\) and \(2x - 3 = 0\), dividing both sides by 2, we get \(x = -2\) and \(x = \frac{3}{2}\).

Key concepts

function transformations

Function transformations involve changing the appearance or position of a graph through various algebraic modifications. The most common transformations include shifting, reflecting, stretching, and compressing. Understanding the basics of these transformations helps in analyzing how a function behaves under different conditions. For example, the function given by the exercise, experiences different transformations such as horizontal stretching and vertical scaling.

finding zeros of functions

To find the zeros of a function, you need to determine the values of x that make the function equal to zero. For the original function in the exercise, \(f(x)=(x+4)(x-3)\), we set f(x) to zero, \( (x+4)(x-3) = 0 \), and solve for x. The solutions are \(x = -4 \) and \(x = 3\), the points where the graph would intersect the x-axis. This method can be applied after any transformation to see how the zeros shift.

horizontal stretching and compression

Horizontal stretching and compression change the width of the graph. When a function is horizontally stretched, the graph becomes wider, whereas horizontal compression makes it narrower. In the exercise, \(y=f\bigg(\frac{1}{2}x\bigg)\) is a horizontal stretch by a factor of 2. Hence, the zeros change from \(-4, 3\) to \(-8, 6\). Similarly, \(y=f(2x)\) compresses the graph horizontally by a factor of 2, changing zeros from \(-4, 3\) to \(-2, \frac{3}{2}\).

vertical scaling

Vertical scaling involves multiplying the function by a constant. It affects the height of the graph but does not change the zeros. For example, in the exercise, \y = 4f(x)\, the graph is vertically scaled by a factor of 4. Even though it stretches the graph, the zeros remain the same \((-4, 3)\). The height of each point on the graph is now four times higher or lower but the locations of the x-intercepts are unchanged.