All Algebra II Resources
Example Questions
Example Question #1 : Graphing Circular Inequalities
Given the above circle inequality, is the shading on the graph inside or outside the circle?
Can't Tell
Inside
Outside
Both
Outside
Check the center of the circle to see if that point satisfies the inequality.
When evaluating the function at the center (1,1), we see that it does not satisfy the equation, so it cannot be in the shaded region of the graph.
Therefore the shading is outside of the circle.
Example Question #1 : Graphing Circular Inequalities
Given the above circle inequality, which point is not on the edge of the circle?
This is a graph of a circle with radius of 6 and a center of (-2,4). The point (2,2) is not on the edge of the circle, so that is the correct answer. All other points are exactly 6 units away from the circle's center, making them a part of the circle.
Example Question #1 : Graphing Circular Inequalities
Given the above circle inequality, which point satisfies the inequality?
The left side of the equation must be greater than or equal to 36 in order to satisfy the equation, so plugging in each of the values for x and y, we see:
Therefore only yields an answer that is greater than or equal to 36.
Example Question #1 : Graphing Circular Inequalities
Given the above circle inequality, does the center satisfy the equation?
Can't tell
No
Maybe
Yes
No
The center of the circle is , so plugging those values in for x and y yields the response,
Therefore, the center does not satisfy the inequality.
Example Question #2 : Graphing Circular Inequalities
Given the above circle inequality, is the shading on the graph inside or outside the circle?
Outside
Inside
Both
Can't tell
Outside
Check the center of the circle to see if that point satisfies the inequality. When evaluating the function at the center (-2,4), we see that it does not satisfy the equation, so it cannot be in the shaded region of the graph. Therefore the shading is outside of the circle.
Example Question #2 : Graphing Circular Inequalities
Given the above circle inequality, which point is not on the edge of the circle?
Recall the equation of a circle:
where r is the radius and (h,k) is the center of the circle.
This is a graph of a circle with radius of 2 and a center of (-4,-3). The point (2,3) is not on the edge of the circle, so that is the correct answer.
All other points are exactly 2 units away from the circle's center, making them a part of the circle's edge.
Example Question #1 : Graphing Circular Inequalities
Given the above circle inequality, which point satisfies the inequality?
The left side of the equation must be less than or equal to 4 in order to satisfy the equation, so plugging in each of the values for x and y, we see:
The only point that satisfies the inequality is the point (-3,-2), since it yields an answer that is less than or equal to 4.
Example Question #11 : Graphing Circular Inequalities
Given the above circle inequality, does the center satisfy the equation?
No
Yes
Can't tell
Maybe
Yes
Recall the equation of circle:
where r is the radius and the center of the circle is at (h,k).
The center of the circle is (-4,-3), so plugging those values in for x and y yields the response that 0 is less than or equal to 4, which is a true statement, so the center does satisfy the inequality.
Example Question #12 : Graphing Circular Inequalities
Given the above circle inequality, is the shading on the graph inside or outside the circle?
Inside
Both
Outside
Can't Tell
Inside
Check the center of the circle to see if that point satisfies the inequality. When evaluating the function at the center (-4,-3), we see that it does satisfy the equation, so it can be in the shaded region of the graph. Therefore the shading is inside of the circle.
Example Question #1053 : Algebra Ii
What is the -intercept of ?
There are no -intercepts of this function.
The -intercepts of a function are the points where . When we substitute this into our equation, we get:
.
Adding nine to both sides,
.
Modifying the equation to get like bases get us,
Since .
Now we can set the exponents equal to eachother and solve for .
Thus,
.
Giving us our final solution:
.
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