Which graph represents the solution set of the system of inequalities?
1 answer:
Answer:
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
(see image)
bottom right image
Explanation:
First try the origin (0,0) to rule out two of the graphs.
3y ≥ x - 9
3(0) ≥ (0) - 9
3 ≥ - 9 yes
3x + y > - 3
3(0) + (0) > - 3
3 > - 3 yes
so the origin should be in the shaded area of the graph, which rules out the top right and bottom left graphs.
Now try a coordinate that is in the shaded area of one of the remaining graphs, but not in the other one. If it works, the graph is the one that has that point in the shaded region, and vice versa.
Try point (4, 2)
3y ≥ x - 9
3(2) ≥ (4) - 9
6 ≥ - 5 yes
3x + y > - 3
3(4) + (2) > - 3
12 + 2 > - 3
14 > - 3 yes
So the graph is the bottom right one since (4, 2) is included in that shaded region.
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Answer:
Answers are in bold
Step-by-step explanation:
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<u>Part 1A</u>
Use the Zero Product Property to find the x-intercepts:
0=-(x+4)(x-1)
0=(x+4)(x-1)
0=(x+4)
-4=x
x=-4
0=(x-1)
1=x
x=1
So the x-intercepts are x=-4 and x=1 which is also (-4,0) and (1,0)
For the y-intercepts, plug in x=0:
f(x)=-(0+4)(0-1)
f(x)=(-4)(-1)
f(x)=4
So the y-intercept is y=4 which is also (0,4)
<u>Part 1B</u>
In general, the axis of symmetry is equal to x=h in the equation y=(x-h)^2+k
For a parabola, the axis of symmetry is equal to x=-b/2a, so we expand the given equation and then plug in the values of "a" and "b" to get the axis of symmetry:
f(x)=-(x+4)(x-1)
f(x)=-(x^2+3x-4)
f(x)=-x^2-3x+4
So now we know a=-1 and b=-3, therefore we can plug them in:
x=-b/2a
x=-(-3)/2(-1)
x=3/-2
x=-3/2
x=-1.5
So the equation of the axis of symmetry is x=-1.5
<u>Part 1C</u>
We'll use the equation x=-b/2a to get the x-coordinate for the vertex (which we already know is x=-1.5), then plug in that value of x in the original equation to get the y-coordinate for the vertex:
f(x)=-(-1.5+4)(-1.5+1)
f(x)=-(2.5)(-0.5)
f(x)=-(-1.25)
f(x)=1.25 <-- y-coordinate
So the coordinates of the vertex are (-1.5,1.25)
<u>Part 1D</u>
See the image below
I'll let someone else do the last 4 questions as I am unable to answer them.
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