Instruction

1

This problem can be solved analytically, therefore, we can not draw graphs of

**straight**and*parabola*. This often gives an advantage in sample solution, as the task can be given such functions, that it is easier and faster to draw.2

According to the textbooks on algebra parabola set function of the form f(x)=ax^2+bx+c, where a,b,c are real numbers, while the coefficient a is different it is zero. The function g(x)=kx+h, where k h is a real number, defines a line in the plane.

3

The point

**of intersection of****straight**and*parabola*is a common point of both curves, so it functions take the same value, that is, f(x)=g(x). This statement allows to write the equation: ax^2+bx+c=kx+h, which will give the opportunity to find many points**of intersection**.4

In the equation ax^2+bx+c=kx+h, you must move all terms to the left side and lead like this: ax^2+(b-k)x+c-h=0. It now remains to solve the obtained quadratic equation.

5

All found the "x" is not the answer to the problem, as a point in the plane is characterized by two real numbers (x,y). To complete the solution, it is necessary to calculate the corresponding "allocated prizes as stated". To do this, substitute the "x" in either the function f(x) or the function g(x), as for the point

**of intersection**is true: y=f(x)=g(x). Then you will find all common points*of the parabola*and**straight**.6

To consolidate the material it is very important to consider the solution to the example. Let the parabola is given by f(x)=x^2-3x+3 and the line – g(x)=2x-3. Write down the equation f(x)=g(x), that is x^2-3x+3=2x-3. Transferring all terms to the left side, and bringing a similar, we get: x^2-5x+6=0. The roots of this quadratic equation: x1=2, x2=3. Now find the corresponding "allocated prizes as stated": y1=g(x1)=1, y2=g(x2)=3. Thus, all points

**of intersection**: (2,1) and (3,3).# Advice 2 : How to find the intersection of a line and parabola

The task of finding the points

**of intersection of**any figures is ideologically simple. The difficulties in them are only for arithmetic, as it allowed various typos and errors.Instruction

1

This problem can be solved analytically, therefore, we can not draw graphs of

**straight**and*parabola*. This often gives an advantage in sample solution, as the task can be given such functions, that it is easier and faster to draw.2

According to the textbooks on algebra parabola set function of the form f(x)=ax^2+bx+c, where a,b,c are real numbers, while the coefficient a is different it is zero. The function g(x)=kx+h, where k h is a real number, defines a line in the plane.

3

The point

**of intersection of****straight**and*parabola*is a common point of both curves, so it functions take the same value, that is, f(x)=g(x). This statement allows to write the equation: ax^2+bx+c=kx+h, which will give the opportunity to find many points**of intersection**.4

In the equation ax^2+bx+c=kx+h, you must move all terms to the left side and lead like this: ax^2+(b-k)x+c-h=0. It now remains to solve the obtained quadratic equation.

5

All found the "x" is not the answer to the problem, as a point in the plane is characterized by two real numbers (x,y). To complete the solution, it is necessary to calculate the corresponding "allocated prizes as stated". To do this, substitute the "x" in either the function f(x) or the function g(x), as for the point

**of intersection**is true: y=f(x)=g(x). Then you will find all common points*of the parabola*and**straight**.6

To consolidate the material it is very important to consider the solution to the example. Let the parabola is given by f(x)=x^2-3x+3 and the line – g(x)=2x-3. Write down the equation f(x)=g(x), that is x^2-3x+3=2x-3. Transferring all terms to the left side, and bringing a similar, we get: x^2-5x+6=0. The roots of this quadratic equation: x1=2, x2=3. Now find the corresponding "allocated prizes as stated": y1=g(x1)=1, y2=g(x2)=3. Thus, all points

**of intersection**: (2,1) and (3,3).# Advice 3 : How to find the coordinates of the point of intersection of two straight lines

If two lines are not parallel, they will intersect at one point. Find

**the coordinates of the****point***of intersection*of two straight lines can be both graphical and computational way, depending on what data provides the task.You will need

- - two lines in the drawing.
- - equation of two straight lines.

Instruction

1

If straight already drawn on the chart, find the solution graphically. This will continue with both or one of the lines so that they intersect. Then mark the intersection point and drop from it perpendicular to the x-axis (usually Oh).

2

With the scale divisions marked on the axis, find the value of x for that point. If it is on the positive axis direction (right from zero), its value will be positive, otherwise – negative.

3

Similarly, find the ordinate of the point of intersection. If the projection point is located above zero – it is positive, below is negative. Write down the coordinates of a point in form (x, y) is the solution of the problem.

4

If direct is specified in the form formula y=KX+b, you can also solve the problem graphically: draw straight on a coordinate grid and find the solution as described above.

5

Try to find a solution to the problem using the given formula. To do this, make these equations system and solve it. If the equation in the form y=KX+b, just Paranaita both parts with x find X. Then substitute the x value into one of the equations and find y.

6

You can find a solution to the method of Kramer. In that case, give the equation to the form А1х+В1у+C1=0 and А2х+В2у+C2=0. According to the Cramer formula x=-(С1В2-С2В1)/(А1В2-А2В1), and y=-(А1С2-А2С1)/(А1В2-А2В1). Please note, if the denominator is zero, then the lines are parallel or coincide, respectively, do not intersect.

7

If you are given straight lines in space in the canonical form, before you start searching for the solutions, check for parallel direct. For this, rate coefficients at t, if they are proportional, such that x=-1+3t, y=7+2t, z=2+t and x=-1+6t, y=-1+4t, z=-5+2t, then the lines are parallel. In addition, direct can interbreed, in this case, the system will have no solutions.

8

If you found out that the lines intersect, find their point of intersection. First Paranaita variables from different direct, conventionally, replacing t on u first and v second-straight. For example, if you direct this x=t-1, y=2t+1, z=t+2 and x=t+1, y=t+1, z=2t+8 you will get an expression of type u-1=v+1, 2u+1=v+1, u+2=2v+8.

9

Express one equation of u, substitute in another and find v (in this problem u=-2,v=-4). Now, to find the point of intersection, substitute the obtained values in place of t (no matter the first or the second equation) and get the coordinates of the point x=-3, y=-3, z=0.