How do you find the vertex and the intercepts for #f(x)=2x^2-12x+21#?

2 Answers
Apr 17, 2017

#v=(6,21)# and there are no interceptions.

Explanation:

The vertex is the minimum of the function, so if we set the standard notation #f(x)=ax^2+bx+c# it is expressed as

#v_x=-b/{2a}=6# and #v_y=f(v_x)=21#

The intercepts are the roots of #f#, but #f# has no real roots since

#Delta/4=(b/2)^2-ac=36-42<0#

Apr 17, 2017

Y-intercept: #(0,21)#
Vertex: #(3,3)#
No x intercepts

Explanation:

#f(x)=2x^2-12x+21#

Y-intercept:
Find #f(0)#:
#f(0)=2(0)^2-12(0)+21=21#

Vertex- Complete the square of the equation
#f(x)=2[x^2-6x+10.5]#

#=2[(x-3)^2+1.5]#

#=2(x-3)^2+3#

Looking at the vertex form, the vertex is at #(3,3)#

X-intercepts:
Methods include using the quadratic formula, factoring, graphing, or completing the square/vertex form.
Using vertex form as found above:

#0=f(x)=2(x-3)^2+3#
#-3/2=(x-3)^2#
#x=+-sqrt(-3/2)+3#

This is a square root of a negative number, meaning the function has no #x# intercepts.