Answer:
(a)(i) g(0) = 4
(ii) g(-1) = 3
(iii) a = -√6 ≈ -2.4
(iv) a = -2
(b) Domain: [-3, 2)
Range: [-5, 4]
Step-by-step explanation:
Part (a)(i)
To find the value of g(0) on the given graph, locate the point where the graph intersects the y-axis (where x = 0) and read the corresponding y-coordinate. Therefore:
[tex]\Large\boxed{\boxed{g(0) = 4}}[/tex]
Part (a)(ii)
To find the value of g(-1) on the given graph, locate the point on the x-axis where x = -1. From that point, trace vertically to the curve of the graphed function, and then read the corresponding y-coordinate. This y-coordinate represents the value of the function g(-1). Therefore:
[tex]\Large\boxed{\boxed{g(-1) = 3}}[/tex]
Part (a)(iii)
The value of 'a' when g(a) = -2 is the x-value of the point where y = -2 on the graph. To find the value of 'a', locate the point on the graph where the function's output is -2 and identity its x-coordinate, which is approximately x ≈ -2.5.
To find the exact value of 'a', we need to substitute y = -2 into the equation of the function g(x). The equation of the function has not been given, however it is g(x) = -x² + 4 with a restricted domain and range. If we substitute g(a) = -2 into the equation and solve for a, we get:
[tex]\begin{aligned}-a^2+4&=-2\\-a^2&=-6\\a^2&=6\\a&=\pm\sqrt{6}\end{aligned}[/tex]
As the only point on the curve where y = -2 is when x is negative, the exact value of 'a' is the negative solution. Therefore:
[tex]\Large\boxed{\boxed{a = -\sqrt{6}}}[/tex]
Part (a)(iv)
The value of 'a' when g(a) = 0 is the x-value of the point where y = 0 on the graph, which corresponds to the x-axis. Therefore, to find the value of 'a', determine the value of x where the function intersects the x-axis.
The function intersects the x-axis at both x = -2 and x = 2, however, the presence of an open circle at (2, 0) indicates that this point is not included in the solution. So, the only point of intersection with the x-axis is (-2, 0). Therefore:
[tex]\Large\boxed{\boxed{a = -2}}[/tex]
Part (b)
The domain of a function is the set of all possible input values (x-values) for which the function is defined. The open circle at the rightmost endpoint of the curve (2, 0) indicates that this point is not included in the domain, and the closed circle at the leftmost endpoint of the curve (-3, -5) indicates that this point is included in the domain. Therefore, the domain of the function is:
[tex]\Large\boxed{\boxed{\textsf{Domain:}\quad [-3, 2)}}[/tex]
The range of a function is the set of all possible output values (y-values) for which the function is defined. The maximum value of the function occurs at the turning point of the curve, specifically at y = 4. The minimum value of the function occurs at the leftmost endpoint of the curve when y = -5. Since this endpoint is marked with a closed circle, it is included in the range. Therefore, the range of the function is:
[tex]\Large\boxed{\boxed{\textsf{Range:}\quad [-5, 4]}}[/tex]
