Question 4 (Fill-In-The-Blank Worth 6 points)
1 answer:
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Answer:
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 <em>x</em> 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
It looks like you might have intended to say the roots are 7 + i and 5 - i, judging by the extra space between 7 and i.
The simplest polynomial with these characteristics would be
but seeing as each of the options appears to be a quartic polynomial, I suspect f(x) is also supposed to have only real coefficients. In that case, we need to pair up any complex root with its conjugate to "complete" f(x). We end up with
which appears to most closely resemble the third option. Upon expanding, we see f(x) does indeed have real coefficients: