Defining figures on the coordinate plane using equations and inequalities. Lesson topic: “An equation with two variables and its graph” Graphing a quadratic equation with two variables

Linear equation with two variables - any equation that has the following form: a*x + b*y =с. Here x and y are two variables, a,b,c are some numbers.

The solution to the linear equation a*x + b*y = c is any pair of numbers (x,y) that satisfies this equation, that is, turns the equation with variables x and y into a correct numerical equality. A linear equation has an infinite number of solutions.

If each pair of numbers that are solutions to a linear equation in two variables is depicted on coordinate plane in the form of points, then all these points form the graph of a linear equation with two variables. The coordinates of the points will be our x and y values. In this case, the x value will be the abscissa, and the y value will be the ordinate.

Graph of a Linear Equation in Two Variables

The graph of a linear equation with two variables is the set of all possible points on the coordinate plane, the coordinates of which will be solutions to this linear equation. It is easy to guess that the graph will be a straight line. That is why such equations are called linear.

Construction algorithm

Algorithm for plotting a linear equation in two variables.

1. Draw coordinate axes, label them and mark the unit scale.

2. In a linear equation, put x = 0, and solve the resulting equation for y. Mark the resulting point on the graph.

3. In a linear equation, take the number 0 as y, and solve the resulting equation for x. Mark the resulting point on the graph

4. If necessary, take an arbitrary value of x and solve the resulting equation for y. Mark the resulting point on the graph.

5. Connect the resulting points and continue the graph beyond them. Sign the resulting straight line.

Example: Graph the equation 3*x - 2*y =6;

Let's put x=0, then - 2*y =6; y= -3;

Let's put y=0, then 3*x = 6; x=2;

We mark the obtained points on the graph, draw a straight line through them and label it. Look at the figure below, the graph should look exactly like this.

Graphing an equation is much easier than many people think. You don't have to be a math genius or first in your math class to understand the basic principles of this process. This article describes how to graph linear and quadratic equations and inequalities, as well as equations with moduli.

Steps

Linear equation graph

Use the formula y=mx+b. To graph a linear equation, you simply plug the values ​​into the formula.

• This formula establishes the relationship between variables x And y.
• Parameter m corresponds to the slope of the line. In other words, m indicates the rate of increase (or decrease) y with change x.
• Parameter b indicates where the line corresponding to the equation intersects the axis y.

1. Make a graph. The linear equation is depicted most simply, since there is no need to calculate anything before plotting the graph. First, construct a rectangular coordinate system.

   Find the point where the line intersects the axis y(This b). For example, in the case of the equation y=2x-1 parameter b is equal to -1, that is, the line intersects the axis y at point -1.

   • At the point of intersection of the axis y coordinate x always takes the value 0. Thus, in our example, the intersection point has coordinates (0,-1).
   • Mark on the graph the point where the line intersects the axis y.

2. Find the slope of the line. For a straight line, the slope corresponds to the parameter m. In the case of the equation y=2x-1 this parameter is equal to 2. However, note that the slope indicates a change y with growth x, that is, it should be represented as a fraction. Since at the coordinate x costs the integer 2, you can write the slope as 2/1.

   • To graph slope, start at the axis intercept y. In this case, the change in coordinates y corresponds to the numerator, and the change in coordinate x- the denominator of the fraction.
   • In our example, you can start from point -1 and move from it up by 2 and to the right by 1.
   • A positive slope means that with growth x you're climbing up y, while with a negative slope y decreases. Variable x increases to the right along the horizontal axis and decreases to the left.
   • When determining the slope, you can use as many points as you like, although one point is sufficient.

3. Draw a straight line. After you determine the slope of the line and plot at least one point, you can connect it to the intersection point of the axis y and draw a straight line. Extend the line to the edges of the graph and draw arrows at the ends to indicate that it continues further.

   Inequality graph with one variable

   1. Draw a number line. Since a single axis is sufficient to depict an inequality with one variable, there is no need to draw a rectangular coordinate system. Instead, just draw a straight line.

      Draw the inequality. This is quite simple since there is only one coordinate. Suppose we need to represent the inequality x<1. Для начала следует найти на оси число 1.

      Draw a line. Draw a line from the point you just marked on the number line. If the variable is greater than this number, move the line to the right. If the variable is smaller, draw a line to the left. Place an arrow at the end of the line to show that it is not a final segment and continues further.

      Check the answer. Substitute for the variable x any number and mark its position on the number line. If this number lies on the line you drew, the graph is correct.

   Linear inequality graph

   Use the straight line formula. A similar formula was used above for ordinary linear equations, but in this case, instead of the ‘=’ sign, you should put an inequality sign. This may be one of the following signs:<, >, ≤ (\displaystyle \leq ) or ≥ (\displaystyle \geq ).

   • The equation of a straight line is y=mx+b, Where m corresponds to the slope, and b- intersection with the axis y.
   • The inequality sign means that this expression has multiple solutions.

   1. Draw the inequality. Find the point of intersection of the line with the axis y and its slope, then mark the corresponding coordinates. As an example, consider the inequality y>1/2x+1. In this case, the straight line will intersect the axis y at x=1, and its slope will be ½, that is, when moving to the right by 2 units, we will rise up by 1 unit.

      Draw a line. Before doing this, look at the inequality sign. If this< или >, you should draw a dotted line. If the inequality contains the sign ≤ (\displaystyle \leq ) or ≥ (\displaystyle \geq ), the line must be solid.

      Shade the graph. Since the inequality has many solutions, the graph should show all possible solutions. This means shading the area above or below the line.

   Quadratic Equation Graph

   Look at the formula. In a quadratic equation, at least one variable is squared. Typically, a quadratic equation is written as follows: y=ax 2 +bx+c.

   • When plotting quadratic equation you will get a parabola, that is, a curve in the form Latin letter'U'.
   • To construct a parabola, you need to know the coordinates of at least three points, including the vertex of the parabola (its central point).

   1. Define a, b And c. For example, in Eq. y=x 2 +2x+1 a=1, b=2 and c=1. Each parameter is a number that precedes the variable to the corresponding power. For example, if before x doesn't cost any number, it means b=1, since the corresponding term can be written in the form 1 x.

      Find the vertex of the parabola. To find the midpoint of a parabola, use the expression -b/2a. For our example, we get -2/2(1), that is -1.

      Make a table. So we know that the coordinate x vertices is equal to -1. However, this is only one coordinate. To find the corresponding coordinate y, as well as the other two points of the parabola, you need to make a table.

      Construct a table of three rows and two columns.

      • Write down the coordinate x the vertices of the parabola in the central cell of the left column.
      • Select two more coordinates x at the same distance to the left and right (in negative and positive side along the horizontal axis). For example, you can move 2 units to the left and right from the vertex, that is, write -3 and 1 in the corresponding cells.
      • You can choose any integers that are equidistant from the vertex.
      • If you want to build a more accurate graph, you can use five points instead of three. In this case, you should do the same, only the table will consist not of three, but of five rows.

   2. Use an equation and table to find unknown coordinates y. Take one x coordinate from the table, substitute it into behind given equation and find the corresponding y coordinate.

      • In our case, we substitute into the equation y=x 2 +2x+1 instead x-3. As a result we find y= -3 2 +2(-3)+1, that is y=4.
      • We write down the found coordinate y in a cell near its corresponding coordinate x.
      • Find all three (or five if you use more points) coordinates this way y.

   3. Plot the points on the graph. So, you now have at least three points with known coordinates that can be marked on the graph. Connect them with a parabola-shaped curve. Ready!

Let it be given equation with two variables F(x; y). You have already become familiar with ways to solve such equations analytically. Many solutions of such equations can be represented in graph form.

The graph of the equation F(x; y) is the set of points on the coordinate plane xOy whose coordinates satisfy the equation.

To graph equations in two variables, first express the y variable in the equation in terms of the x variable.

Surely you already know how to build various graphs of equations with two variables: ax + b = c – straight line, yx = k – hyperbola, (x – a) 2 + (y – b) 2 = R 2 – circle whose radius is equal to R, and the center is at point O(a; b).

Example 1.

Graph the equation x 2 – 9y 2 = 0.

Solution.

Let's factorize the left side of the equation.

(x – 3y)(x+ 3y) = 0, that is, y = x/3 or y = -x/3.

Answer: Figure 1.

A special place is occupied by defining figures on a plane with equations containing the sign of the absolute value, which we will dwell on in detail. Let's consider the stages of constructing graphs of equations of the form |y| = f(x) and |y| = |f(x)|.

The first equation is equivalent to the system

(f(x) ≥ 0,
(y = f(x) or y = -f(x).

That is, its graph consists of graphs of two functions: y = f(x) and y = -f(x), where f(x) ≥ 0.

To plot the second equation, plot two functions: y = f(x) and y = -f(x).

Example 2.

Graph the equation |y| = 2 + x.

Solution.

The given equation is equivalent to the system

(x + 2 ≥ 0,
(y = x + 2 or y = -x – 2.

We build many points.

Answer: Figure 2.

Example 3.

Plot the equation |y – x| = 1.

Solution.

If y ≥ x, then y = x + 1, if y ≤ x, then y = x – 1.

Answer: Figure 3.

When constructing graphs of equations containing a variable under the modulus sign, it is convenient and rational to use area method, based on dividing the coordinate plane into parts in which each submodular expression retains its sign.

Example 4.

Graph the equation x + |x| + y + |y| = 2.

Solution.

In this example, the sign of each submodular expression depends on the coordinate quadrant.

1) In the first coordinate quarter x ≥ 0 and y ≥ 0. After expanding the module, the given equation will look like:

2x + 2y = 2, and after simplification x + y = 1.

2) In the second quarter, where x< 0, а y ≥ 0, уравнение будет иметь вид: 0 + 2y = 2 или y = 1.

3) In the third quarter x< 0, y < 0 будем иметь: x – x + y – y = 2. Перепишем этот результат в виде уравнения 0 · x + 0 · y = 2.

4) In the fourth quarter, when x ≥ 0, and y< 0 получим, что x = 1.

We will plot this equation by quarters.

Answer: Figure 4.

Example 5.

Draw a set of points whose coordinates satisfy the equality |x – 1| + |y – 1| = 1.

Solution.

The zeros of the submodular expressions x = 1 and y = 1 divide the coordinate plane into four regions. Let's break down the modules by region. Let's arrange this in the form of a table.

Region	Submodular expression sign	The resulting equation after expanding the module
I	x ≥ 1 and y ≥ 1	x + y = 3
II	x< 1 и y ≥ 1	-x + y = 1
III	x< 1 и y < 1	x + y = 1
IV	x ≥ 1 and y< 1	x – y = 1

Answer: Figure 5.

On the coordinate plane, figures can be specified and inequalities.

Inequality graph with two variables is the set of all points of the coordinate plane whose coordinates are solutions to this inequality.

Let's consider algorithm for constructing a model for solving inequalities with two variables:

1. Write down the equation corresponding to the inequality.
2. Graph the equation from step 1.
3. Select an arbitrary point in one of the half-planes. Check whether the coordinates of the selected point satisfy this inequality.
4. Draw graphically the set of all solutions to the inequality.

Let us first consider the inequality ax + bx + c > 0. The equation ax + bx + c = 0 defines a straight line dividing the plane into two half-planes. In each of them, the function f(x) = ax + bx + c retains its sign. To determine this sign, it is enough to take any point belonging to the half-plane and calculate the value of the function at this point. If the sign of the function coincides with the sign of the inequality, then this half-plane will be the solution to the inequality.

Let's look at examples graphic solution the most common inequalities with two variables.

1) ax + bx + c ≥ 0. Figure 6.

2) |x| ≤ a, a > 0. Figure 7.

3) x 2 + y 2 ≤ a, a > 0. Figure 8.

4) y ≥ x 2 . Figure 9.

5) xy ≤ 1. Figure 10.

If you have questions or want to practice drawing on a plane model the sets of all solutions to inequalities in two variables using mathematical modeling, you can conduct free 25-minute lesson with an online tutor after you register. To continue working with a teacher, you will have the opportunity to choose a tariff plan that suits you.

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OBJECTIVE:1) To introduce students to the concept of “equation with two variables”;

2) Learn to determine the degree of an equation with two variables;

3) Teach to identify by given function which figure is a graph

given equation;

4) Consider transformations of graphs with two variables;

given equation with two variables using the Agrapher program;

6) Develop logical thinking students.

I. New material - an explanatory lecture with elements of conversation.

(the lecture is conducted using the author’s slides; graphs are drawn in the Agrapher program)

T: When studying lines, two problems arise:

Using the geometric properties of a given line, find its equation;

Inverse problem: given the equation of a line, study its geometric properties.

We considered the first problem in the geometry course in relation to circles and straight lines.

Today we will consider the inverse problem.

Consider equations of the form:

A) x(x-y)=4; b) 2u-x 2 =-2 ; V) x(x+y 2 ) = x +1.

are examples of equations with two variables.

Equations with two variables X And at looks like f(x,y)=(x,y), Where f And – expressions with variables X And u.

If in Eq. x(x-y)=4 substitute in place of variable X its value is -1, and instead at– value 3, then the correct equality will be obtained: 1*(-1-3)=4,

Pair (-1; 3) variable values X And at is a solution to the equation x(x-y)=4.

That is solving the equation with two variables is called the set of ordered pairs of values ​​of variables that form this equation into a true equality.

Equations with two variables usually have infinitely many solutions. Exceptions form, for example, equations such as X 2 +(y 2 - 4) 2 = 0 or

2x 2 + at 2 = 0 .

The first of them has two solutions (0; -2) and (0; 2), the second has one solution (0; 0).

The equation x 4 + y 4 +3 = 0 has no solutions at all. It is of interest when the values ​​of the variables in the equation are integers. By solving such equations with two variables, pairs of integers are found. In such cases, the equation is said to be solved in integers.

Two equations having the same set of solutions are called equivalent equations. For example, the equation x(x + y 2) = x + 1 is an equation of the third degree, since it can be transformed into the equation xy 2 + x 2 - x-1 = 0, the right side of which is a polynomial of the standard form of the third degree.

The degree of an equation with two variables, represented in the form F(x, y) = 0, where F(x, y) is a polynomial of standard form, is called the degree of the polynomial F(x, y).

If all solutions to an equation with two variables are depicted as points in the coordinate plane, you will get a graph of an equation with two variables.

Schedule equation with two variables is the set of points whose coordinates serve as solutions to this equation.

So, the graph of the equation ax + by + c = 0 is a straight line if at least one of the coefficients a or b not equal to zero (Fig. 1). If a = b = c = 0, then the graph of this equation is coordinate plane (Fig. 2), if a = b = 0, A c0, then the graph is empty set (Fig. 3).

Equation graph y = a x 2 + by + c is a parabola (Fig. 4), a graph of the equation xy=k (k0) – hyperbole (Fig. 5). Equation graph X 2 + y 2 = r, where x and y are variables, r is a positive number, is circle with center at the origin and radius equal to r(Fig. 6). The graph of the equation is ellipse, Where a And b– major and minor semi-axes of the ellipse (Fig. 7).

The construction of graphs of some equations is facilitated by the use of their transformations. Let's consider converting graphs of equations in two variables and formulate the rules by which the simplest transformations of equation graphs are performed

1) The graph of the equation F (-x, y) = 0 is obtained from the graph of the equation F (x, y) = 0 using symmetry about the axis u.

2) The graph of the equation F (x, -y) = 0 is obtained from the graph of the equation F (x, y) = 0 using symmetry about the axis X.

3) The graph of the equation F (-x, -y) = 0 is obtained from the graph of the equation F (x, y) = 0 using central symmetry about the origin.

4) The graph of the equation F (x-a, y) = 0 is obtained from the graph of the equation F (x, y) = 0 by moving parallel to the x-axis by |a| units (to the right, if a> 0, and to the left if A < 0).

5) The graph of the equation F (x, y-b) = 0 is obtained from the graph of the equation F (x, y) = 0 by moving to |b| units parallel to the axis at(up if b> 0, and down if b < 0).

6) The graph of the equation F (ax, y) = 0 is obtained from the graph of the equation F (x, y) = 0 by compressing to the y-axis and a times, if A> 1, and by stretching from the y-axis by times, if 0< A < 1.

7) The graph of the equation F (x, by) = 0 is obtained from the graph of the equation F (x, y) = 0 using compression to the x-axis in b times if b> 1, and by stretching from the x axis by times if 0 < b < 1.

If the graph of some equation is rotated by a certain angle near the origin, then the new graph will be the graph of another equation. The special cases of rotation at angles of 90 0 and 45 0 are important.

8) The graph of the equation F (x, y) = 0 as a result of a clockwise rotation near the origin of coordinates by an angle of 90 0 turns into the graph of the equation F (-y, x) = 0, and counterclockwise into the graph of the equation F (y , -x) = 0.

9) The graph of the equation F (x, y) = 0 as a result of a clockwise rotation near the origin of coordinates by an angle of 45 0 turns into the graph of the equation F = 0, and counterclockwise into the graph of the equation F = 0.

From the rules we have considered for transforming graphs of equations with two variables, rules for transforming graphs of functions are easily obtained.

Example 1. Let us show that by graphing the equation X 2 + y 2 + 2x – 8y + 8 = 0 is a circle (Fig. 17).

Let's transform the equation as follows:

1) group the terms containing the variable X and containing a variable at, and imagine each group of terms in the form of a complete square trinomial: (x 2 + 2x + 1) + (y 2 -2*4*y + 16) + 8 – 1 – 16 = 0;

2) write the resulting trinomials as the square of the sum (difference) of two expressions: (x + 1) 2 + (y – 4) 2 - 9 = 0;

3) let’s analyze, according to the rules for transforming graphs of equations with two variables, the equation (x + 1) 2 + (y – 4) 2 = 3 2: the graph of this equation is a circle with a center at the point (-1; 4) and a radius of 3 units .

Example 2: Let's graph the equation X 2 + 4у 2 = 9 .

Let's imagine 4y 2 in the form (2y) 2, we get the equation x 2 + (2y) 2 = 9, the graph of which can be obtained from the circle x 2 + y 2 = 9 by compressing the x axis by a factor of 2.

Draw a circle with a center at the origin and a radius of 3 units.

Let's reduce the distance of each point from the X axis by 2 times and get a graph of the equation

x 2 + (2y) 2 = 9.

We obtained the figure by compressing the circle to one of its diameters (to the diameter that lies on the X axis). This figure is called an ellipse (Fig. 18).

Example 3. Let's find out what the graph of the equation x 2 - y 2 = 8 is.

Let's use the formula F= 0.

Substituting in this equation instead of X and instead of Y, we get:

T: What is the graph of the equation y = ?

D: The graph of the equation y = is a hyperbola.

U: We transformed the equation of the form x 2 - y 2 = 8 into the equation y =.

Which line will be the graph of this equation?

D: So, the graph of the equation x 2 - y 2 = 8 is a hyperbola.

U: Which lines are asymptotes of the hyperbola y =.

D: The asymptotes of the hyperbola y = are the straight lines y = 0 and x = 0.

U: When the rotation is completed, these straight lines will turn into straight lines = 0 and = 0, that is, into straight lines y = x and y = - x. (Fig. 19).

Example 4: Let's find out what form the equation y = x 2 of the parabola will take when rotated around the origin by an angle of 90 0 clockwise.

Using the formula F (-y; x) = 0, in the equation y = x 2 we replace the variable x with – y, and the variable y with x. We obtain the equation x = (-y) 2, i.e. x = y 2 (Fig. 20).

We looked at examples of graphs of second-degree equations with two variables and found out that the graphs of such equations can be a parabola, a hyperbola, an ellipse (in particular, a circle). In addition, the graph of an equation of the second degree can be a pair of lines (intersecting or parallel). This is the so-called degenerate case. So the graph of the equation x 2 - y 2 = 0 is a pair of intersecting lines (Fig. 21a), and the graph of the equation x 2 - 5x + 6 + 0y = 0 is parallel lines.

II Consolidation.

(students are given “Instruction cards” for constructing graphs of equations with two variables in the Agrapher program (Appendix 2) and “Practical task” cards (Appendix 3) with the formulation of tasks 1-8. The teacher demonstrates graphs of equations for tasks 4-5 on slides ).

Exercise 1. Which of the pairs (5;4), (1;0), (-5;-4) and (-1; -) are solutions to the equation:

a) x 2 - y 2 = 0, b) x 3 - 1 = x 2 y + 6y?

Solution:

Substituting the coordinates of these points into the given equation, we are convinced that not a single given pair is a solution to the equation x 2 - y 2 = 0, and the solutions to the equation x 3 - 1 = x 2 y + 6y are the pairs (5;4), ( 1;0) and (-1; -).

125 - 1 = 100 + 24 (I)

1 - 1= 0 + 0 (I)

125 – 1 = -100 – 24 (L)

1 – 1 = - - (I)

Answer: A); b) (5;4), (1; 0), (-1; -).

Task 2. Find solutions to the equation xy 2 - x 2 y = 12 in which the value X equals 3.

Solution: 1) Substitute the value 3 instead of X in the given equation.

2) We obtain a quadratic equation for the variable Y, which has the form:

3y 2 - 9y = 12.

4) Let's solve this equation:

3y 2 - 9y – 12 = 0

D = 81 + 144 = 225

Answer: pairs (3;4) and (3;-1) are solutions to the equation xy 2 - x 2 y = 12

Task 3. Determine the degree of the equation:

a) 2y 2 - 3x 3 + 4x = 2; c) (3 x 2 + x)(4x - y 2) = x;

b) 5y 2 - 3y 2 x 2 + 2x 3 = 0; d) (2y - x 2) 2 = x(x 2 + 4xy + 1).

Answer: a) 3; b) 5; at 4; d) 4.

Task 4. Which figure is the graph of the equation:

a) 2x = 5 + 3y; b) 6 x 2 - 5x = y – 1; c) 2(x + 1) = x 2 - y;

d) (x - 1.5)(x – 4) = 0; e) xy – 1.2 = 0; e) x 2 + y 2 = 9.

Task 5. Write an equation whose graph is symmetrical to the graph of the equation x 2 - xy + 3 = 0 (Fig. 24) with respect to: a) axis X; b) axes at; c) straight line y = x; d) straight line y = -x.

Task6. Make up an equation, the graph of which is obtained by stretching the graph of the equation y = x 2 -3 (Fig. 25):

a) from the x-axis 2 times; b) from the y-axis 3 times.

Check with the Agrapher program that the task was completed correctly.

Answer: a)y - x 2 + 3 = 0 (Fig. 25a); b) y-(x) 2 + 3 = 0 (Fig. 25b).

b) the lines are parallel, moving parallel to the x-axis 1 unit to the right and parallel to the y-axis 3 units down (Fig. 26b);

c) straight lines intersect, symmetrical display relative to the x axis (Fig. 26c);

d) straight lines intersect, symmetrical display relative to the y-axis (Fig. 26d);

e) the lines are parallel, symmetrical display relative to the origin (Fig. 26e);

e) straight lines intersect, rotation around the origin by 90 clockwise and symmetrical display relative to the x axis (Fig. 26f).

III. Independent work educational in nature.

(students are given cards “Independent work” and a “Reporting table of the results of independent work”, in which students write down their answers and, after self-testing, evaluate the work according to the proposed scheme) Appendix 4 ..

I. option.

a) 5x 3 -3x 2 y 2 + 8 = 0; b) (x + y + 1) 2 - (x-y) 2 = 2(x + y).

a) x 3 + y 3 -5x 2 = 0; b) x 4 +4x 3 y +6x 2 y 2 + 4xy 3 + y 4 = 1.

x 4 + y 4 -8x 2 + 16 = 0.

a) (x + 1) 2 + (y-1) 2 = 4;

b) x 2 -y 2 = 1;

c) x - y 2 = 9.

x 2 - 2x + y 2 - 4y = 20.

Specify the coordinates of the circle's center and its radius.

6. How should the hyperbola y = be moved on the coordinate plane so that its equation takes the form x 2 - y 2 = 16?

Check your answer by graphing using Agrapher.

7. How should the parabola y = x 2 be moved on the coordinate plane so that its equation takes the form x = y 2 - 1

Option II.

1. Determine the degree of the equation:

a)3xy = (y-x 3)(x 2 +y); b) 2y 3 + 5x 2 y 2 - 7 = 0.

2. Is the pair of numbers (-2;3) a solution to the equation:

a) x 2 -y 2 -3x = 1; b) 8x 3 + 12x 2 y + 6xy 2 + y 3 = -1.

3. Find the set of solutions to the equation:

x 2 + y 2 -2x – 8y + 17 = 0.

4. What kind of curve (hyperbola, circle, parabola) is a set of points if the equation of this curve has the form:

a) (x-2) 2 + (y + 2) 2 =9

b) y 2 - x 2 =1

c) x = y 2 - 1.

(check with the Agrapher program that the task was completed correctly)

5. Using the Agrapher program, plot the equation:

x 2 + y 2 - 6x + 10y = 2.

6. How should the hyperbola y = be moved on the coordinate plane so that its equation takes the form x 2 - y 2 = 28?

7. How should the parabola y = x 2 be moved on the coordinate plane so that its equation takes the form x = y 2 + 9.

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