SOLUTION: Given the polynomial 3x^2-6x-18 what is the vertex of it's graph

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Given the polynomial 3x^2-6x-18 what is the vertex of it's graph      Log On


   



Question 316554: Given the polynomial 3x^2-6x-18 what is the vertex of it's graph
Found 2 solutions by Fombitz, josmiceli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square to convert to vertex form,
y=a%28x-h%29%5E2%2Bk
where (h,k) is the vertex.
y=3x%5E2-6x-18
y=3%28x%5E2-2x%29-18
y=3%28x%5E2-2x%2B1%29-18-3
y=3%28x-1%29%5E2-21
.
.
.
(1,-21) is the vertex.
.
.
.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The vertex is midway between the 2 roots
When the equation is in the form
ax%5E2+%2B+bx+%2B+c+=+0
The formula to find roots is
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Midway between the 2 roots is x=+%28-b%29%2F%282a%29
3x%5E2+-+6x+-+18+=+0
The x coordinate of the vertex is at %28-%28-6%29%29%2F%282%2A3%29+=+1
y+=+3%2A1%5E2+-+6%2A1+-+18
y+=+3+-+6+-+18
y+=+-21
The vertex is at (1,-21)
Here's the plot
+graph%28+500%2C+500%2C+-10%2C+10%2C+-25%2C+10%2C+3x%5E2+-+6x+-+18%29+