# Math 22 Graph of Equation

## The Graph of an Equation

The graph of an equation is the set of all points that are solutions of the equation.

In this section, we use point-plotting method. With this method, you construct a table of values that consists of several solution points of the equation

For example, sketch the graph of ${\displaystyle y=2x+1}$. We can construct the table below by plugging points for ${\displaystyle x}$.

 x 0 1 2 3 y=2x+1 1 3 5 7

So, we can sketch the graph from those order pairs.

## Intercepts of a Graph

Some solution points have zero as either the ${\displaystyle x}$-coordinate or the ${\displaystyle y}$-coordinate. These points are called intercepts because they are the points at which the graph intersects the ${\displaystyle x}$- or ${\displaystyle y}$-axis.

 To find ${\displaystyle x}$-intercepts, let ${\displaystyle y}$ be zero and solve the equation for ${\displaystyle x}$.

To find ${\displaystyle y}$-intercepts, let ${\displaystyle x}$ be zero and solve the equation for ${\displaystyle y}$.


Example Find the x-intercepts and y-intercepts of the function ${\displaystyle y=x^{2}-2x}$

Solution:
x-intercept: Let ${\displaystyle y=0}$, so ${\displaystyle x^{2}-2x=0}$, hence ${\displaystyle x(x-2)=0}$, therefore, ${\displaystyle x=0}$ or ${\displaystyle x=2}$
y-intercept: Let ${\displaystyle x=0}$, so ${\displaystyle y=(0)^{2}-2(0)=0}$
Answer: ${\displaystyle (0,0)}$ and ${\displaystyle (2,0)}$ are x-intercepts
${\displaystyle (0,0)}$ is y-intercept

## Circles

 The standard form of the equation of a circle is

${\displaystyle (x-h)^{2}+(y-k)^{2}=r^{2}}$

The point ${\displaystyle (h,k)}$ is the center of the circle, and the positive number ${\displaystyle r}$ is the radius of the circle


In general, to write an equation of a circle, we need to know radius ${\displaystyle r}$ and the center ${\displaystyle (h,k)}$.

Example Given that the point ${\displaystyle (2,1)}$ is on the circle centered at (3,4). Find the equation of a circle.

Solution:
We need to know the radius and the center in order to write the equation. The center is given at ${\displaystyle (3,4)}$. It is left to find the radius.
Radius is the distance between the center and a point on the circle. So, radius ${\displaystyle r}$ is the distance between ${\displaystyle (2,1)}$ and ${\displaystyle (3,4)}$.
So, ${\displaystyle r={\sqrt {(2-3)^{2}+(1-4)^{2}}}={\sqrt {1+9}}={\sqrt {10}}}$
Now, write the equation of the circle with radius ${\displaystyle r={\sqrt {10}}}$ and center ${\displaystyle (3,4)}$ to get:
${\displaystyle (x-3)^{2}+(y-4)^{2}=10}$

## Notes

Distance ${\displaystyle D}$ between ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$ can be calculated by using ${\displaystyle D={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}}$

## Points of Intersection

An ordered pair that is a solution of two different equations is called a point of intersection of the graphs of the two equations

For example, find the point(s) of intersection of two equations ${\displaystyle y=2x-1}$ and ${\displaystyle y=3x+4}$.

The order pairs that satisfy both of these equation should have the same ${\displaystyle y}$ value, so ${\displaystyle y=2x-1=3x+4}$

Then, ${\displaystyle 2x-1=3x+4}$

Therefore ${\displaystyle x=-5}$