Difference between revisions of "Math 22 Graph of Equation"

Latest revision as of 06:50, 19 July 2020

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 $y=2x+1$ . We can construct the table below by plugging points for $x$ .

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 $x$ -coordinate or the $y$ -coordinate. These points are called intercepts because they are the points at which the graph intersects the $x$ - or $y$ -axis.

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

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

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

Circles

 The standard form of the equation of a circle is
 
  $(x-h)^{2}+(y-k)^{2}=r^{2}$ 
 
 The point  $(h,k)$  is the center of the circle, and the positive number  $r$  is the radius of the circle

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

Example Given that the point $(2,1)$ 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 $(3,4)$ . It is left to find the radius.
Radius is the distance between the center and a point on the circle. So, radius $r$ is the distance between $(2,1)$ and $(3,4)$ .
So, $r={\sqrt {(2-3)^{2}+(1-4)^{2}}}={\sqrt {1+9}}={\sqrt {10}}$
Now, write the equation of the circle with radius $r={\sqrt {10}}$ and center $(3,4)$ to get:
$(x-3)^{2}+(y-4)^{2}=10$

Notes

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

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 $y=2x-1$ and $y=3x+4$ .

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

Then, $2x-1=3x+4$

Therefore $x=-5$

Return to Topics Page

This page were made by Tri Phan

@@ Line 3: / Line 3: @@
 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 <math>y=2x+1</math>. We can construct the table below by plugging points for <math>x</math>.
+{|class = "wikitable" style = "text-align:center"
+| [[x]]
+| [[0]]
+| [[1]]
+| [[2]]
+| [[3]]
+|-
+| [[y=2x+1]]
+| [[1]]
+| [[3]]
+| [[5]]
+| [[7]]
+|}
+So, we can sketch the graph from those order pairs.
+[[File:graph 1.2.png|center|500px]]
 ==Intercepts of a Graph==
 Some solution points have zero as either the <math>x</math>-coordinate or the <math>y</math>-coordinate. These points are called intercepts because they are the points at which the graph intersects the <math>x</math>- or <math>y</math>-axis.
+  To find <math>x</math>-intercepts, let <math>y</math> be zero and solve the equation for <math>x</math>.
+  To find <math>y</math>-intercepts, let <math>x</math> be zero and solve the equation for <math>y</math>.
+'''Example''' Find the x-intercepts and y-intercepts of the function <math>y=x^2-2x</math>
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
+|-
+|'''x-intercept''': Let <math>y=0</math>, so <math>x^2-2x=0</math>, hence <math>x(x-2)=0</math>, therefore, <math>x=0</math> or <math>x=2</math>
+|-
+|'''y-intercept''': Let <math>x=0</math>, so <math>y=(0)^2-2(0)=0</math>
+|-
+|'''Answer''': <math>(0,0)</math> and <math>(2,0)</math> are x-intercepts
+|-
+|<span style="display:inline-block; width: 54px;"></span> <math>(0,0)</math> is y-intercept
+|}
+==Circles==
+  The standard form of the equation of a circle is
+  <math>(x-h)^2+(y-k)^2=r^2</math>
+  The point <math>(h,k)</math> is the center of the circle, and the positive number <math>r</math> is the radius of the circle
+In general, to write an equation of a circle, we need to know radius <math>r</math> and the center <math>(h,k)</math>.
+'''Example''' Given that the point <math>(2,1)</math> is on the circle centered at (3,4). Find the equation of a circle.
+{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
+!Solution: &nbsp;
+|-
+|We need to know the radius and the center in order to write the equation. The center is given at <math>(3,4)</math>. It is left to find the radius.
+|-
+|Radius is the distance between the center and a point on the circle. So, radius <math>r</math> is the distance between <math>(2,1)</math> and <math>(3,4)</math>.
+|-
+|So, <math>r=\sqrt{(2-3)^2+(1-4)^2}=\sqrt{1+9}=\sqrt{10}</math>
+|-
+|Now, write the equation of the circle with radius <math>r=\sqrt{10}</math> and center <math>(3,4)</math> to get:
+|-
+|<math>(x-3)^2+(y-4)^2=10</math>
+|}
+==Notes==
+Distance <math>D</math> between <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math> can be calculated by using <math>D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}</math>
+==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 <math>y=2x-1</math> and <math>y=3x+4</math>.
+The order pairs that satisfy both of these equation should have the same <math>y</math> value, so <math>y=2x-1=3x+4</math>
+Then, <math>2x-1=3x+4</math>
+Therefore <math>x=-5</math>
+[[Math_22| '''Return to Topics Page''']]
 '''This page were made by [[Contributors|Tri Phan]]'''