Question 395569: find the vertex, the line of symmetry, and the maximim or minimum value of the quadratic function, and graph the function on paper.
f(x) = 2x^2+2x+4
what is the x-coordinate of the vertex? (type a simplified fraction)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the vertex, the line of symmetry, and the maximim or minimum value of the quadratic function, and graph the function on paper.
f(x) = 2x^2+2x+4
what is the x-coordinate of the vertex? (type a simplified fraction)
Just by inspection of the equation, you should know that f(x) is a parbola that opens upwards because the coefficient of the x^2 term is positive, therefore, it has a minimum. Conversely, if the coefficient is negative, the parabola will open downward and therefore, have a maximum.
There are 2 methods for finding the coordinates of the vertex.
The first method uses the formula -b/2a, a, being the coefficient of the x^2 term, and,b, being the coefficient of the x term. This formula gives you the x coordinate of the vertex. Substitute this value into the equation to get the y coordinate.
The second method is to complete the square and put it into this form:
y=(x-h)^2+k,(h,k) being the (x,y) coordinates of the vertex.
Solve y=2x^2+2x+4 using the first method
a=2
b=2
-b/2a=-2/4=-1/2
x=-1/2
y=2(-1/2)^2+2(-1/2)+4
=1/2-1+4=3+1/2
(x,y) coordinates of the vertex are (-1/2,3+1/2)
3+1/2 is the minimum value, and the line of symmetry is x=-1/2
Solve y=2x^2+2x+4 using the second method (completing the square)
first, factor out 2 from the x^2 and x terms, but leave the constant, 4, outside
2(x^2+x+ )+4
complete the square by taking half of the coefficient of the x term then squaring it.
2(x^2+x+1/4)-1/2+4 (since we added 2*1/4 to the equation we must subtract an equal amount = 1/2)
=2(x+1/2)^2+3+1/2
This is now in the form that shows the (x,y) coordinates are (-1/2,3+1/2), same as found by the first method.
ans: coordinates of vertex are (-1/2,3+1/2)
maximum value = 3+1/2
line of symmetry = x=-1/2
See below the graph of parabola
|
|
|
