Question 189058: I have to find the vertex but I don't really understand the steps to it. f(x)=x^2+10x+14
do I have to do x^2 +10x+14-14?
the other one is f(x)=-x^2 +6x+20 to find the vertex and to determine if there is a maximum or minimum value and what that value is...
the max.value would be k which would be 20 right? and minimum?
Anna
Found 2 solutions by Tutor44, solver91311:
Answer by Tutor44(1)   (Show Source):
You can put this solution on YOUR website! To find the vertex and examine if there is a maximum or minimum value you have to differentiate that function and put it equal to zero as following, I'll take the second problem in consideration, f(x) = - x^2 +6x +20
d/dx{f(x)} = d/dx(-x^2)+ d/dx (6x) + d/dx(20)
=-2x + 6 + 0
=-2x+6
now set, d/dx {f(x)} = 0
or, -2x+6=0
or, x= 3 put this in f(x), so y=f(3)=-3^2+6*3+20=29
so the vertex is (3,29)=(x,y)
Explanation: Differentiating with respect to (w.r.t) x means the rate of change of the value of the function w.r.t small change in value of x. so that point will be the highest (or lowest) where the change is zero. So we set it equal to zero.
Maximum or Minimum value: now at this point, the question is whether at this point the function has its highest value(maximum) or lowest value(minimum). If we again differentiate f(x) then if the differentiated value is positive (that means the value of the rate of change functions is increasing so it will be a minima) but if it is negative it means that the rate is decreasing indicating the value is maximum.
so d/dx^2 {f(x)}
= d/dx {d/dx {f(x)}
= d/dx (-2x) + d/dx(6)
= -2+0
= -2
so that at this point the function has a maximum value
and the value is, f(3) = 29
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
The vertex of a parabola is a point on the plane, so you have to determine the x and y coordinates in order to define that point. Fortunately, we have a formula available for just this purpose.
The x-coordinate of the vertex of a parabola whose equation is in the form:
 = ax^2 + bx + c)
is given by:
For your given function,
 = x^2 + 10x + 14)
a = 1 and b = 10, so:
} = -5)
Now that we know the x-coordinate of the vertex, we can find the value of the function at that x value to compute the y-coordinate of the vertex, thus:
\ =\ f(-5)\ =\ (-5)^2 + 10(-5) + 14\ =\ 25 - 50 + 14\ =\ -11)
So, the vertex is the point (-5,-11)
For your other problem, perform the same process to discover the coordinates of the vertex. You will find that the value of the function at the vertex (same thing as the y-coordinate of the vertex), given properly performed arithmetic, is something other than 20. As to whether it is a maximum or minimum, you can tell by the sign on the lead coefficient.
If the lead coefficient is > 0, then the parabola opens upward and the value of the function at the vertex is a minimum. If the lead coefficient is < 0, then it is the opposite case, i.e the graph opens downward and the vertex is a maximum. And, of course, if the lead coefficient = 0, then you don't have a parabola at all, rather you have a straight line.
For your second problem, the lead coefficient, the a in  = ax^2 + bx + c) , is -1, so you have a graph that opens downward and the vertex is a maximum.
John
|
|
|
