SOLUTION: Hello what is the vertex of this quadratic and can you show me the steps of how you got this: f(x)= -3/10x^2+9/2x-7/6 thank you

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Question 553590: Hello what is the vertex of this quadratic and can you show me the steps of how you got this:
f(x)= -3/10x^2+9/2x-7/6
thank you
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
what is the vertex of this quadratic and can you show me the steps of how you got this:
f(x)= (-3/10)x^2+(9/2)x-(7/6)
--------
vertex occurs at x = -b/(2a)
= (-9/2)/(2(-3/10))
= (-9/2)/(-3/5)
= (-45/-6)
----
= 15/2
----
= 7.5
=============
f(7.5) = (-3/10)(7.5)^2+(9/2)(7.5)-(7/6) = 15.71
------
Vertex: (7.5 , 15.71)
===========================
Cheers,
Stan H.
===========================

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The x-coordinate of the vertex is at
+x+=+-b%2F%282a%29+ where the equation
has the form +f%28x%29+=+ax%5E2+%2B+b%2Ax+%2B+c+|
--------
+f%28x%29+=+%28-3%2F10%29%2Ax%5E2+%2B+%289%2F2%29%2Ax+-+7%2F6+
+a+=+-3%2F10+
+b+=+9%2F2+
+-b%2F%282a%29+=+-%289%2F2%29%2F%282%2A%28-3%2F10%29%29+
+-b%2F%282a%29+=+%28-10%2F6%29%2A%28-9%2F2%29+
+-b%2F%282a%29+=+90%2F12+
+-b%2F%282a%29+=+15%2F2+
----------------
So far, the vertex is at ( 15/2,y )
Now find the y-coordinate
+f%28+15%2F2+%29+=+%28-3%2F10%29%2A%2815%2F2%29%5E2+%2B+%289%2F2%29%2A%2815%2F2%29+-+7%2F6+
+f%28+15%2F2+%29+=+%28-3%2F10%29%2A%28225%2F4%29+%2B+135%2F4+-+7%2F6+
+f%28+15%2F2+%29+=+-675%2F40+%2B+540%2F40+-+7%2F6+
+f%2815%2F2%29+=+-2025%2F120+%2B+1620%2F120+-+140%2F120+
+f%2815%2F2%29+=+-545%2F120+
+f%2815%2F2%29+=+-109%2F24+
The vertex is at (15/2, -109/24)
Unless I made a mistake