SOLUTION: Find the minimum y-value located on the graph of y=3x^2+2x-5. This is what I have 3x^2+2x-5 3(X^2+2x)-5 3(x^2+1x+5-5)-5 3(x^2+1+5)-1-5 3(x+1)^2 The vortex is (-1,-6) am I co

Algebra ->  Equations -> SOLUTION: Find the minimum y-value located on the graph of y=3x^2+2x-5. This is what I have 3x^2+2x-5 3(X^2+2x)-5 3(x^2+1x+5-5)-5 3(x^2+1+5)-1-5 3(x+1)^2 The vortex is (-1,-6) am I co      Log On


   

Question 947491: Find the minimum y-value located on the graph of y=3x^2+2x-5. This is what I have
3x^2+2x-5
3(X^2+2x)-5
3(x^2+1x+5-5)-5
3(x^2+1+5)-1-5
3(x+1)^2
The vortex is (-1,-6) am I correct?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
3x%5E2%2B2x-5
3%28x%5E2%2B%282%2F3%29x%29-5....complete square
3%28x%5E2%2B%282%2F3%29x%2B1%2F9%29-3%2A1%2F9-5
3%28x%2B1%2F3%29%5E2-1%2F3-15%2F3
3%28x%2B1%2F3%29%5E2-16%2F3

since
h=-1%2F3
k=-16%2F3
the vertex is at (-1%2F3,-16%2F3)

the minimum y-value located on the graph is y=-16%2F3