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.