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How to find the coordinates of the points of intersection with the axes. Coordinates of the intersection point of function graphs. The case of two nonlinear functions

In practice and in textbooks, the most common methods listed below are for finding the intersection point of various function graphs.

First way

The first and simplest is to take advantage of the fact that at this point the coordinates will be equal and equate the graphs, and from what you get you can find $x$. Then substitute the found $x$ into any of the two equations and find the coordinate of the game.

Example 1

Let's find the intersection point of two lines $y=5x + 3$ and $y=x-2$, equating the functions:

$x=-\frac(1)(2)$

Now let’s substitute the x we ​​received into any graph, for example, choose the one that is simpler - $y=x-2$:

$y=-\frac(1)(2) – 2 = - 2\frac12$.

The intersection point will be $(-\frac(1)(2);- 2\frac12)$.

Second way

The second method is that a system is compiled from existing equations, by means of transformations one of the coordinates is made explicit, that is, expressed through the other. After this expression in the given form is substituted into another.

Example 2

Find out at what points the graphs of the parabola $y=2x^2-2x-1$ and the straight line $y=x+1$ intersect.

Solution:

Let's create a system:

$\begin(cases) y=2x^2-2x-1 \\ y= x + 1 \\ \end(cases)$

The second equation is simpler than the first, so let’s substitute it for $y$:

$x+1 = 2x^2 – 2x-1$;

$2x^2 – 3x – 2 = 0$.

Let’s calculate what x is equal to, to do this we will find the roots that make the equality true, and write down the answers received:

$x_1=2; x_2 = -\frac(1)(2)$

Let’s substitute our results along the x-axis one by one into the second equation of the system:

$y_1= 2 + 1 = 3; y_2=1 - \frac(1)(2) = \frac(1)(2)$.

The intersection points will be $(2;3)$ and $(-\frac(1)(2); \frac(1)(2))$.

Third way

Let's move on to the third method - graphical, but keep in mind that the result it gives is not quite accurate.

To apply the method, both function graphs are plotted on the same scale on the same drawing, and then a visual search for the intersection point is performed.

This method is good only if an approximate result is sufficient, and also if there is no data on the patterns of the dependencies under consideration.

  • To find the coordinates of the intersection point of the graphs of functions, you need to equate both functions to each other, move all terms containing $ x $ to the left side, and the rest to the right side, and find the roots of the resulting equation.
  • The second method is to create a system of equations and solve it by substituting one function into another
  • The third method involves graphically constructing functions and visually determining the intersection point.
  • The case of two linear functions

    Consider two linear functions $ f(x) = k_1 x+m_1 $ and $ g(x) = k_2 x + m_2 $. These functions are called direct. It is quite easy to construct them; you need to take any two values ​​$ x_1 $ and $ x_2 $ and find $ f(x_1) $ and $ (x_2) $. Then repeat the same with the function $ g(x) $. Next, visually find the coordinate of the intersection point of the function graphs.

    You should know that linear functions have only one intersection point and only when $ k_1 \neq k_2 $. Otherwise, in the case $ k_1=k_2 $ the functions are parallel to each other, since $ k $ is the slope coefficient. If $ k_1 \neq k_2 $, but $ m_1=m_2 $, then the intersection point will be $ M(0;m) $. It is advisable to remember this rule to quickly solve problems.

    Example 1
    Let $ f(x) = 2x-5 $ and $ g(x)=x+3 $ be given. Find the coordinates of the intersection point of the function graphs.
    Solution

    How to do it? Since two linear functions are presented, the first thing we look at is the slope coefficient of both functions $ k_1 = 2 $ and $ k_2 = 1 $. We note that $ k_1 \neq k_2 $, so there is one intersection point. Let's find it using the equation $ f(x)=g(x) $:

    $$ 2x-5 = x+3 $$

    We move the terms with $ x $ to the left side, and the rest to the right:

    $$ 2x - x = 3+5 $$

    We have obtained $ x=8 $ the abscissa of the intersection point of the graphs, and now let’s find the ordinate. To do this, let’s substitute $ x = 8 $ into any of the equations, either in $ f(x) $ or in $ g(x) $:

    $$ f(8) = 2\cdot 8 - 5 = 16 - 5 = 11 $$

    So, $ M (8;11) $ is the point of intersection of the graphs of two linear functions.

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    Answer
    $$ M (8;11) $$
    The case of two nonlinear functions
    Example 3
    Find the coordinates of the intersection point of the function graphs: $ f(x)=x^2-2x+1 $ and $ g(x)=x^2+1 $
    Solution

    What about two nonlinear functions? The algorithm is simple: we equate the equations to each other and find the roots:

    $$ x^2-2x+1=x^2+1 $$

    We distribute terms with and without $ x $ on different sides of the equation:

    $$ x^2-2x-x^2=1-1 $$

    The abscissa of the desired point has been found, but it is not enough. The ordinate $y$ is still missing. We substitute $ x = 0 $ into any of the two equations of the problem condition. For example:

    $$ f(0)=0^2-2\cdot 0 + 1 = 1 $$

    $ M (0;1) $ - intersection point of function graphs

    Answer
    $$ M (0;1) $$

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