Question

Consider the graph below.

Answer 1

Answer:

$f(x)=\begin{cases}x^2 + 3; & \text{ } x$

Step-by-step explanation:

The y-intercept (when x = 0) on the quadratic graph is 3, the upper quadratic graph does not intersect the x-axis, and the endpoint of the function has an open circle that stops at x = 1. The starting point extends to the boundaries of the graph

Therefore, the piecewise function shown in the quadratic graph is given as follows;

f(x) = x² + 3; x < 1

The straight line graph has a negative slope and starts at x = 1, with a closed circle, extending to higher values of x to the boundaries of the graph

Two points on the straight line graph are (1, 3) and (5, -5), therefore, the slope is (-5 - 3)/(5 - 1) = -2

The equation of the function is y - 3 = -2·(x - 1)

∴ y = f(x)  -2·x + 2 + 3 = -2·x + 5

f(x) = -2·x + 5; x ≥ 1