How do you graph #f(x)=x^2+x+1#?

Question

Julia Adams · Accepted Answer

Part 1 of 2 - General description of processes
See part 2 of 2 for actual calculations

You will need to determine the important points before creating a table of values. #color(blue)("Point 1 - general shape of the graph")# The coefficient of #x^2# is #+1# (positive) so the graph is of general shape #uu# thus it has a minimum. Just for a moment let is pretend that the coefficient was negative. If that had been the case the graph would be of general shape #nn# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Point 2 - What are the coordinates of the minimum (vertex)")# #color(brown)("method 1")# You can use the formula #x=(-b+-sqrt(b^2-4ac))/(2a)# to determine the x-intercepts. Find the mid point and that is the #x# coordinate for the minimum. Substitute back into the equation to determine the #y# coordinate. #color(brown)("method 2 - not very well known or used")# Start the process of completing the square (you do not need to do all of it). #color(brown)("method 3")# Complete the square which 'almost' gives you the coordinates directly. The #y# can be read off directly but you need to multiply the #x# equivalent by (-1)# #y=ax^2+bx+c color(white)("d")->y=a(x+b/(2a))^2+k+c# #k# is needed as a correction as changing the original equation this way introduces an error. The inclusion of #k# neutralises that error. #y_("vertex")=k+c# #x_("vertex")=(-1)xx(b/(2a))^2# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Point 3- determine the y-intercept")# Set the value of #x# as 0. Substitute and solve for y Shortcut #-> # it is the value of #c# in #y=ax^2+bx+c# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Point 4- determine the x-intercept")# Sometimes you can just factorise and set #y=0 # Use the above if you can. It is much less work Set the value of #y# to 0 and solve for #x# using: #y=a(x+b/(2a))^2+k+c# or #x=(-b+-sqrt(b^2-4ac))/(2a)#

Genesis Allen · Answer

Solution 2 of 2 - The actual calculations

The graph is of general shape #uu# as the #x^2# term is positive
Given: #y=ax^2+bx+c color(white)("d")->color(white)("d")x^2+x+1#

#color(blue)("Determine the y-intercept")#
#y_("intercept")=c=+1 -> (x,y)=(0,1)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex")#

You can not factorise. If we try #(x+1)(x+1)# we get #x^2+2x+1#

#color(brown)("Sort of cheat method"))#

write in form #y=a(x^2+b/ax)+c color(white)("d")->color(white)("d")y=1xx[x^2+(color(red)(1)/1xx x)]+c#

#color(white)(dddddddddddddddddd"d")=>color(white)("d")x_("vertex")=(-1/2)xxcolor(red)(1)/1= color(green)(-1/2)#

#y=x^2+x+1color(white)("d")->color(white)("d")y_("vertex")=(color(green)(-1/2))^2+(color(green)(-1/2))+1#

#color(white)("ddddddddddddd")->color(white)("d")y_("vertex")=+1/4-1/2+1=+3/4#

Vertex #color(white)("d")->color(white)("d")(x,y)=(-1/2,3/4)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the x-intercept")#

Note that the graph is of form #uu# and that #y_("vertex")=+3/4#

So there is no graph below #y=+3/4#. Thus the graph can not cross the x-axis. Thus there is no x-intercept for what is called real numbers. There is if you use complex numbers.

#ax+bx+c => a=1; b=1; c=1#

#x=(-b+-sqrt(b^2-4ac))/(2a) color(white)("d")->color(white)("d")x= (-1+-sqrt(1^2-4(1)(1)))/(2(1))#

#color(white)("dddddddddddddddddd")->color(white)("d")x=(-1+-sqrt(3xx(-1)))/2#

#color(white)("dddddddddddddddddd")->color(white)("d")x=(-1+-sqrt(3)color(white)("d")i)/2#

Where #i# is the square root of negative 1

Eli Assouline · Answer

undefined To graph the function $ f(x) = x^2 + x + 1 $, you can follow these steps:

1. Determine the vertex of the parabola using the formula $ x = -\frac{b}{2a} $. In this case, $ a = 1 $ and $ b = 1 $, so the vertex occurs at $ x = -\frac{1}{2(1)} = -\frac{1}{2} $.

2. Find the corresponding $ y $-coordinate for the vertex by substituting $ x = -\frac{1}{2} $ into the function. $ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{5}{4} $.

3. Plot the vertex at $ \left(-\frac{1}{2}, \frac{5}{4}\right) $.

4. Determine additional points by choosing other $ x $-values. For example, you can choose $ x = -1, 0, 1 $ to calculate corresponding $ y $-values.

- $ f(-1) = (-1)^2 - 1 + 1 = 1 $
- $ f(0) = 0^2 + 0 + 1 = 1 $
- $ f(1) = 1^2 + 1 + 1 = 3 $

5. Plot these points on the graph: $ (-1, 1) $, $ (0, 1) $, and $ (1, 3) $.

6. Draw a smooth curve through the plotted points to represent the graph of $ f(x) = x^2 + x + 1 $.

The graph of the function will be a parabola opening upwards, with the vertex at $ \left(-\frac{1}{2}, \frac{5}{4}\right) $ and passing through the points $ (-1, 1) $, $ (0, 1) $, and $ (1, 3) $.