Question

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

Answer 1

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

Explanation:

You can build a table of values but you will need to identify the key points.

`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`
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`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`
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`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`
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`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)`

Answer 2

Solution 2 of 2 - The actual calculations

Explanation:

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)`

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`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)`
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`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

Tony B