Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 150921: Sketch the graph of f(x)= x2-3 for f(2): Sketch the graph of f(x)= x2-3 for f(2)
Answer by jim_thompson5910(8735) About Me  (Show Source):
You can put this solution on YOUR website!

Table of Contents:


Jump to Table
Jump to Graph


In order to graph y=x^2-3, we need to plot some points.


Let's find y when x=-3
(Note: you can start with any x-value):


y=x^2-3 Start with the given equation.


y=(-3)^2-3 Plug in x=-3.


y=1(9)-3 Square -3 to get 9.


y=9-3 Multiply 1 and 9 to get 9.


y=6 Combine like terms.


So when x=-3, then y=6.


So we have the point .


----------------------------------------------------


Let's find y when x=-2:


y=x^2-3 Start with the given equation.


y=(-2)^2-3 Plug in x=-2.


y=1(4)-3 Square -2 to get 4.


y=4-3 Multiply 1 and 4 to get 4.


y=1 Combine like terms.


So when x=-2, then y=1.


So we have the point .


----------------------------------------------------


Let's find y when x=-1:


y=x^2-3 Start with the given equation.


y=(-1)^2-3 Plug in x=-1.


y=1(1)-3 Square -1 to get 1.


y=1-3 Multiply 1 and 1 to get 1.


y=-2 Combine like terms.


So when x=-1, then y=-2.


So we have the point .


----------------------------------------------------


Let's find y when x=0:


y=x^2-3 Start with the given equation.


y=(0)^2-3 Plug in x=0.


y=1(0)-3 Square 0 to get 0.


y=0-3 Multiply 1 and 0 to get 0.


y=-3 Combine like terms.


So when x=0, then y=-3.


So we have the point .


----------------------------------------------------


Let's find y when x=1:


y=x^2-3 Start with the given equation.


y=(1)^2-3 Plug in x=1.


y=1(1)-3 Square 1 to get 1.


y=1-3 Multiply 1 and 1 to get 1.


y=-2 Combine like terms.


So when x=1, then y=-2.


So we have the point .


----------------------------------------------------


Let's find y when x=2:


y=x^2-3 Start with the given equation.


y=(2)^2-3 Plug in x=2.


y=1(4)-3 Square 2 to get 4.


y=4-3 Multiply 1 and 4 to get 4.


y=1 Combine like terms.


So when x=2, then y=1.


So we have the point .


----------------------------------------------------


Let's find y when x=3:


y=x^2-3 Start with the given equation.


y=(3)^2-3 Plug in x=3.


y=1(9)-3 Square 3 to get 9.


y=9-3 Multiply 1 and 9 to get 9.


y=6 Combine like terms.


So when x=3, then y=6.


So we have the point .


----------------------------------------------------





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Now let's make a table of the values we just found.


Table of Values:


xy
-36
-21
-1-2
0-3
1-2
21
36

Now let's plot the points:


 drawing(900, 900, -10, 10, -10, 10,<BR>
grid(1),<BR>
graph(900, 900, -10, 10, -10, 10, 0),<BR>
circle(-3,6,0.08),circle(-3,6,0.10),<BR>
circle(-2,1,0.08),circle(-2,1,0.10),<BR>
circle(-1,-2,0.08),circle(-1,-2,0.10),<BR>
circle(0,-3,0.08),circle(0,-3,0.10),<BR>
circle(1,-2,0.08),circle(1,-2,0.10),<BR>
circle(2,1,0.08),circle(2,1,0.10),<BR>
circle(3,6,0.08),circle(3,6,0.10)
<BR>

)




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Graph:


Now draw a curve through all of the points to graph y=x^2-3:


 drawing(900, 900, -10, 10, -10, 10,<BR>
grid(1),<BR>
graph(900, 900, -10, 10, -10, 10, x^2-3),<BR>
circle(-3,6,0.08),circle(-3,6,0.10),<BR>
circle(-2,1,0.08),circle(-2,1,0.10),<BR>
circle(-1,-2,0.08),circle(-1,-2,0.10),<BR>
circle(0,-3,0.08),circle(0,-3,0.10),<BR>
circle(1,-2,0.08),circle(1,-2,0.10),<BR>
circle(2,1,0.08),circle(2,1,0.10),<BR>
circle(3,6,0.08),circle(3,6,0.10)
<BR>

) Graph of y=x^2-3