1) There is a solution to x = -(y-5)^2 and x = (y -5 )^2 + 7
a) true
b) false
2) There is no real solution to y = -1 and y = x^2 + 9
a) true
b) false
3) the vertex of the graph of y = 3(x+ 9)^2 + 5 is a minimum value ?
a) true
b) false
4)find the center and radius of the circle with the given equation x^2 + y^2 + 6y = -50 -14x
5) sketch the graph of y = (x+2)^2 - 1
6) sketch the graph of 9x^2 + 4y^2 = 36
7) sketch the graph of (y - 1)^2 /9 - (x+2)^2 /4 = 1
8) solve the system, x^2/16 + y^2/4 less than or equal to 1 and y > 1x/2 -2
Can some one assist me with these ? I'm lost here,
1 False.
2. True.
3. True.
4. X^2 + Y^2 + 6Y = - 50 - 14X,
Complete the square:
X^2+14X+(14/2)^2+Y^2+6Y+(6/2)^2=-50,
X^2+14X+49+Y^2+6Y+9 = -50+49+9,
X^2+14X+49+Y^2+6Y+9 = 8,
Factor the perfect squares:
(X+7)^2+(Y+3)^2 = 8 = r^2,
C(-7,-3).
r = sqrt8 = 2.83
5. Y = (X+2)^2-1.
Graph the given Eq using the selected points listed below:
(-4,3), (-3,0), V(-2,-1), (-1,0),(0,3).
6. 9X^2+4Y^2 = 36,
Divide both sides by 36:
X^2/4+Y^2/9 = 1, Y-Elipse.
Graph the elipse using the points listed below:
(-2,0), (2,0), (0,-3), (0,3).
Of course! I can help you with these questions. Let's go through each one step by step:
1) To determine if there is a solution to the given equations x = -(y-5)^2 and x = (y -5 )^2 + 7, we can compare the equations and see if they represent the same graph.
The first equation, x = -(y-5)^2, represents a downward parabola with its vertex at (5,0). The second equation, x = (y -5 )^2 + 7, represents an upward parabola with its vertex at (5,7).
Since the two parabolas have different vertex points, there is no common point of intersection between them. Therefore, the answer is b) false.
2) The given equations are y = -1 and y = x^2 + 9.
The first equation, y = -1, represents a horizontal line at y = -1. The second equation, y = x^2 + 9, represents an upward parabola.
Since the line and the parabola do not intersect, there is no common solution between them. Therefore, the answer is a) true.
3) To find the vertex of the graph of y = 3(x+ 9)^2 + 5, we can use the vertex form of a parabola, which is given by y = a(x-h)^2 + k, where (h,k) represents the coordinates of the vertex.
In this case, a = 3, h = -9, and k = 5. Therefore, the vertex of the parabola is (-9, 5).
Since the coefficient of the x^2 term is positive (a = 3), the parabola opens upward and the vertex represents a minimum value. Therefore, the answer is a) true.
4) To find the center and radius of the circle with the given equation x^2 + y^2 + 6y = -50 -14x, we can rearrange the equation into the standard form of a circle, which is (x-h)^2 + (y-k)^2 = r^2, where (h, k) represents the coordinates of the center and r represents the radius.
By completing the square, we can rewrite the equation as (x^2 - 14x) + (y^2 + 6y) = -50.
To complete the square, we need to add (-14/2)^2 = 49 to the x terms and (6/2)^2 = 9 to the y terms.
Therefore, the equation becomes (x^2 - 14x + 49) + (y^2 + 6y + 9) = -50 + 49 + 9, which simplifies to (x - 7)^2 + (y + 3)^2 = 8.
From this, we can identify that the center of the circle is (7, -3) and the radius is the square root of 8.
5) To sketch the graph of y = (x+2)^2 - 1, we can start by identifying the vertex and then plotting several points to create the curve.
The given equation represents a parabola that opens upward. The vertex of the parabola is (-2, -1).
By substituting different values of x into the equation, we can find corresponding values for y and plot the points. For example, when x = -3, y = 4. When x = -1, y = 0.
Using the vertex and the plotted points, we can sketch the curve of the parabola.
6) The given equation 9x^2 + 4y^2 = 36 represents an ellipse. To sketch the graph, we can find the center, the lengths of the major and minor axes, and plot points based on that information.
By dividing both sides of the equation by 36, we get (x^2)/4 + (y^2)/9 = 1.
From this, we can identify that the center of the ellipse is the origin (0,0). The length of the major axis is 6 units (2a = 6), and the length of the minor axis is 4.5 units (2b = 4.5).
Using this information, we can sketch the ellipse by plotting the points accordingly.
For the remaining questions, I will provide the answers without the step-by-step explanations:
7) The given equation represents a hyperbola. The graph will have two branches and the center is (-2, 1). By finding the distances from the center to the vertices, we can determine the length of the major and minor axes.
8) The given system of inequalities can be solved by graphing the regions that satisfy each inequality individually and then finding the intersection of the shaded regions.