How do you graph the parabola #g(x) = x^2 - 4x + 2# using vertex, intercepts and additional points?

2 Answers
Aug 7, 2016

While we're in standard form, let's find the y intercept. This is found by making #x = 0#:

#y = 0^2 - 4(0) + 2 = 2#

Hence, the y intercept is at #y = 2#, or #(0, 2)#.

Now, let's convert to vertex form by completing the square.

#y = x^2 - 4x + 2#

#y = 1(x^2 - 4x + n - n) + 2#

#n = (b/2)^2#

#n = (-4/2)^2#

#n = 4#

#y = 1(x^2 - 4x + 4 - 4) + 2#

#y = 1(x ^2 - 4x + 4) - 4 + 2#

#y = 1(x^2 - 4x + 4) - 2#

#y = 1(x - 2)^2 - 2#

In vertex form, #y = a(x - p)^2 + q#, the vertex is located at #(p, q)#. Hence, the vertex will be at #(2, -2)#.

As for the x intercepts, we have to make #y = 0# and solve:

#0 = 1(x - 2)^2 - 2#

#2 = (x - 2)^2#

#+-sqrt(2) = x - 2#

#x = +-sqrt(2) + 2#

Hence, there will be x intercepts as #(2 + sqrt(2), 0)# and #(2 - sqrt(2), 0)#.

Finally, before graphing, you may find it helpful to prepare a table of values so that you can graph more accurately. It may look like the following.

Let's assume that you're working on the interval #[-1, 3]#:

enter image source here

You can obviously use a larger or a smaller table of values, it just depends on what interval of the function your teacher wants you to graph.

Here is the graph of the function:

enter image source here

Practice exercises:

#1.# Graph the following functions, identifying the vertex, intercepts, and if necessary preparing a table of values.

a) #y = -2x^2#

b) #y = 3x^2 - 2#

c) #y = -4x^2 + 5x#

d) #y = 2x^2 + 4x - 1#

e) #y = -1/4x^2 - 3x + 5#

Hopefully this helps, and good luck!

Aug 7, 2016

Demonstrating a different approach to find the vertex

#"Vertex "->(x,y)=(2,2))#
As the coefficient of #x^2# is positive then the graph of of general shape #uu#. Thus the vertex is a minimum ( the only one #-># Absolute Minimum)

#y_("intercept")=2#

#x_("intercept")=2+sqrt(2)" and "2-sqrt(2)" " larr" exact values"#

#x_("intercept")~~3.41 and x~~0.59" to 2 decimal places"#

Explanation:

As it is not specified how you determine the vertex there is an alternative approach which part way to completing the square.

Consider the general case #y=ax^2+bx+c#

Write this as #y=a(x^2+b/ax)+c#

The #x_("vertex")=(-1/2)xxb/a#

Then by substitution determine #y_("vertex")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)(" Determine the vertex")#

Given:#" "y=x^2-4x+2#

In this case #a=1=> b/a=-4/1=-4#

#=> x_("vertex")=(-1/2)xx(-4)=+2#

#=>y_("vertex") = (4)^2-4(4)+2= 2#

#color(red)("Vertex "->(x,y)=(2,2))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "y_("intercept"))#

Reading directly off the constant for #y=x^2-4x+2#

#color(red)(y_("intercept")=2)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "x_("intercepts")#
There are three options

  1. Factorise - Not always easy to spot the values!
  2. Use the formula
  3. Completing the square.

AS the other solution uses completing the square I will use the formula method. Well worth committing to memory. I did this by making sure I wrote it out each time I practised a question. It has stayed with me for life!

general case #y=ax^2+bx+c#

where #x=(-b+-sqrt(b^2-4ac))/(2a)#

where #a=1; b=-4; c=2# giving:

#x=(-(-4)+-sqrt((-4)^2-4(1)(2)))/(2(1))#

#x=(4+-sqrt(16-8))/2" "=" "2+-(2sqrt(2))/2#

#color(red)(x=2+sqrt(2)" and "x=2-sqrt(2)" " larr" exact values")#

#color(red)(x~~3.41 and x~~0.59" to 2 decimal places")#

Tony B