Answer:
Question 7
Domain: (-2,3]
Range: (1,4]
Question 8
a) No
b) x=4 and x=-2
c) (0,3)
d) (3,0) and (-1,0)
Explanation:
Question 7
Domain refers to the span of x-values that a function encompasses. We can see from the graph that the lowest x value is -2 and the highest one is 3, and we also note the open circle at x=-2 (meaning x=-2 itself isn't included, which we note using curved brackets) and the closed circle at x=3 (meaning x=3 itself is included, which we note using square brackets). This leaves us with a domain of (-2,3]
Range meanwhile refers to the span of y-values that a function encompasses. We can see from the graph that the lowest y value is 1 and the highest one is 4, and we also note the open circle at y=1 (meaning y=1 itself isn't included, which we note using curved brackets) and the closed circle at y=4 (meaning y=4 itself is included, which we note using square brackets). This leaves us with a range of (1,4]
Question 8
a)
If the point (1,-3) is on the graph, then subbing 1 in for x and -3 in for f(x) should equate:
f(x)=x²-2x-3
-3=1²-2(1)-3?
-3=1-2-3?
-3=-4?
Since this doesn't equate, we can say the point is not on the line.
b)
The condition is that (x,5) is on the graph, so let's sub 5 in for f(x):
f(x)=x²-2x-3
5=x²-2x-3
x²-2x-8=0
(x-4)(x+2)=0
x=4 and x=-2
c)
The y-intercept is the point(s) at which the graph crosses the y-axis, ie where x=0, so to find it we sub 0 in for x:
f(x)=x²-2x-3
f(0)=0²-2(0)-3=-3
Therefore the y-intercept is (0,3)
d)
The c-intercept is the point(s) at which the graph crosses the x-axis, ie where y=0, so to find it we sub 0 in for f(x):
f(x)=x²-2x-3
0=x²-2x-3
(x-3)(x+1)=0
x=3 and x=-1
Therefore the x-intercepts are (3,0) and (-1,0)