SOLUTION: Find the vertex of the graph of f(x)= 3x^2-12x+5 and determine whether the graph opens up or down is there a formula to figure this out. please help and thank you

Algebra ->  Linear-equations -> SOLUTION: Find the vertex of the graph of f(x)= 3x^2-12x+5 and determine whether the graph opens up or down is there a formula to figure this out. please help and thank you      Log On


   

Question 1129793: Find the vertex of the graph of f(x)= 3x^2-12x+5 and determine whether the graph opens up or down

is there a formula to figure this out. please help and thank you
Found 3 solutions by stanbon, MathLover1, MathTherapy:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex of the graph of f(x)= 3x^2-12x+5 and determine whether the graph opens up or down
is there a formula to figure this out. please help and thank you
------
The form is y = ax^2+bx+c
a = 3 ; b = -12 ; c = 5
------
If a is positive the graph opens up
If a is negative the graph opens down
-----------------
Where is the vertex ?
The roots are -b/(2a)-sqrt(b^2-4ac) and -b/(2a)+sqrt(b^2-4ac)
----
Notice that -b/(2a) is midway between those roots.
i.e. if the roots are at x = 2 and x = 6, -b/(2a) will be 4
----
The vertex always splits the quadratic into two symmetric wings.
So the vertex must pass thru -b/2a.
-----
If the quadratic goes up or down the vertex is at (-b/(2a),f(-b/(2a))
-----------------
Your problem:: The vertex is at (12/6,f(2))
-----------
Cheers,
Stan H.
----------

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=+ax%5E2%2Bbx%2Bc
When you have a quadratic function in either form, OR , if a+%3E+0, then the parabola opens up , if a+%3C+0, then the parabola opens down .
The vertex is the lowest or highest+point (depending on direction) on the graph of a quadratic function.

f%28x%29=+3x%5E2-12x%2B5 -> a=3->a+%3E+0 and the parabola opens up
to find the vertex of the graph, write your equation in vertex form
The vertex formula is derived from the completing-the-square
f%28x%29=+%283x%5E2-12x%29%2B5
f%28x%29=+3%28x%5E2-4x%2Bb%5E2%29-3b%5E2%2B5
f%28x%29=+3%28x-2%29%5E2-3%2A2%5E2%2B5
f%28x%29=+3%28x-2%29%5E2-12%2B5
f%28x%29=+3%28x-2%29%5E2-7

=> the vertex (h,+k) is (-2,+-7)

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+3%28x-2%29%5E2-7%29+



Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex of the graph of f(x)= 3x^2-12x+5 and determine whether the graph opens up or down

is there a formula to figure this out. please help and thank you
Standard form of a quadratic function: matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+ax%5E2+%2B+bx+%2B+c%29

In this case, a = 3; b = - 12; c = 5

The formula for the x-coordinate or axis of symmetry is: matrix%281%2C3%2C+x%2C+%22=%22%2C+-+b%2F%282a%29%29

 ----- Substituting - 12 for b, and 3 for a

 ------- Substituting 2 for x

highlight_green%28matrix%281%2C3%2C+%22Vertex%3A%22%2C+%22%282%2C%22%2C+%22-+7%29%22%29%29

Since a = 3, a > 0, or is POSITIVE so the graph opens UPWARD.

IGNORE that woman's answer as she posted a WRONG vertex!!