SOLUTION: f(x)=1/5(x+3)^2+3 Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the fuction. Please I need someones assistance.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: f(x)=1/5(x+3)^2+3 Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the fuction. Please I need someones assistance.      Log On


   

Question 353084: f(x)=1/5(x+3)^2+3 Find the vertex, the line of symmetry, and the maximum or minimum value of f(x). Graph the fuction. Please I need someones assistance.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%281%2F5%29%28x%2B3%29%5E2%2B3
For an equation for a parabola in this form:
f%28x%29+=+%281%2F%284p%29%29%28x-h%29%5E2+%2B+k
the vertex is at the point (h, k). All we need to do to your equation to make it match this pattern is to change (x+3) into an equivalent subtraction. A subtraction that is equivalent to (x+3) would be (x-(-3)):
f%28x%29=%281%2F5%29%28x-%28-3%29%29%5E2%2B3
Now we can read the vertex: (-3, 3).
Since the x variable is in the squared expression this is a vertically oriented parabola. A vertically oriented parabola has a vertical line of symmetry and this line passes through the vertex. A vertical line has an equation of the form:
x = some number
and since it must pass through the vertex the line of symmetry for this parabola is:
x = -3
With a positive number in front of the squared expression, this vertically oriented parabola opens upward. And a parabola tha opens upward has a minimum value and no maximum value. The minimum value for the function will be at the vertex and, since the y coordinate of a point is the function's value for that x the minimum value for f(x) is 3.
To graph the function you can use the value of "p". Or you can just create a table of values and plot the points. To create your table of values I suggest using the line of symmetry to advantage. For example:
Let's find the point for x = -2 (which is one larger than the x coordinate of the vertex).
f%28-2%29+=+%281%2F5%29%28%28-2%29%2B3%29%5E2+%2B+3
f%28-2%29+=+%281%2F5%29%281%29%5E2+%2B+3
f%28-2%29+=+%281%2F5%29%281%29+%2B+3
f%28-2%29+=+%281%2F5%29+%2B+3
f%28-2%29+=+3%261%2F5
Now for x = -4 (one less than the x of the vertex), because of the symmetry,
f%28-4%29+=+3%261%2F5 also.
So finding points like this is a two-for-one deal. FInd one point and you automatically know the coordinates of the symmetric point. We can proceed with the points with x's which are 2 more/less than the x of the vertex, then 3 more/less than the x of the vertex, etc. until we have enough points to graph the function.