How do you find the vertex and intercepts for #y = x^2 - 4x#?

2 Answers
Jul 6, 2017

See below.

Explanation:

Quick answer:

#x#-intercepts:
Solve #x^2-4x = 0#

#x(x-4) = 0#,

#x = 0,4#

#y#-intercept:
When #x = 0#, the equation gives us #y = 0#

Vertex: is midway between the intercepts. The midpoint is the average.

#x = (0+4)/2 = 2#

When #x = 2#, the equation gives us #y = (2)^2-4(2) = 4-8=-4#

The vertex is #(2,-4)#

Jul 6, 2017

A different method that uses #color(red)("part of")# completing the square. Purely given to show that sometimes different methods work.

Explanation:

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

#color(blue)("Determine y-intercept")#

Write as #y=x^2-4x+0#
#" "color(red)(uarr)#
#" "color(red)("y-intercept")#

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

Note that in the standardised form of #y=ax^2+bx+c# our equation is such that #a=1# so #b/a=(-4)/1=-4#

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

By substitution

#y_("vertex")=(2)^2-4(2)" "=" "4-8" "=" "-8#

Vertex#->(x,y)=(2,-8)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine x-intercept")#

The #x^2# term is positive so the graph is of form #uu# thus has a minimum. #y_("vertex")<0# so x-intercepts exist.

Factorising

The common term is #x# so lets factor that out giving.

#y=x(x-4)#

The x-intercepts are at #y=0# so by substitution

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

Thus #x=0 and 4#