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Find the vertx, the line of symmetry, the maximum or minimum value of the quadratic...
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For a quadratic equation f(x) = ax^2 + bx + c, the vertex occurs at x = -b/2a; therefore y = f(-b/2a) at the vertex. So, for your equation:
x = -2/(2*2) = -1/2
y = f(-1/2) = 2(-1/2)^2 + 2(-1/2) + 1 = 1/2
For a parabola, the line of symetry passes through the vertex. Therefore the line is x = -1/2.
To know whether the point (-1/2,1/2) is a minimum or maximum either 1) plug in another value for x, like x=0 --> y = 1 > 1/2. Or 2) note that since a > 0, this parabola points up like a bowel. Therefore the vertex is a minimum.
To graph this function, draw 1/2 of the parabola with vertex (-1/2,1/2) and through the point (0,1). then draw the mirror image on the other side of the line of symmetry.
Posted by kjcdb8er on August 19, 2009 at 7:57 AM (Answer #1)
High School Teacher
The second degree equation of the form y= ax^2+bx+c represents a parabola.
y/2 =2(x+1/2)^2+ 1/2. (1),
is maximum as x-> infinity. It is minimum when x+1/2=0 or x=-1/2 and the minimum value of y = 1/2 We can re write (1) as:
(x+1/2)^2 = y/2-1/2= (1/2)(y-1)
or X^2 = 4a Y , where X = x+1/2 an Y = y-1 and a=1/8.
Compare this with the standard form of parabola in analytical geometry Y^2=4aX, which is symmetric about Y=0 or the X axis, with vertex at (X,Y) =(0 , 0) , its focus at (a , 0) and Latus rectum = 4a
The equation at (2), therfore, is a parabola , with line of symmetry x+1/2 = 0 or x=-1/2 , a line parallel to y axis at a distance of -1/2 units from it(y axis).
Tracing of the curve y =2x^2+2x+1 or its graph:
i) y= 2x^2+2x+1 .This is a parabola.
(ii)Symmetry: The parabola is symmetric about x=-1/2
(iii) Its vertex is determined by the equations:
x+1/2 =0 and y-1 =0
or at (x,y) = (-1/2 , 1)
(iv)Its Focus: X = 0 and Y= a or x+1/2=0 and y-1=1/8 or y=1+1/8= 9/8. Threfore, the coordinates of the focus :(x,y) = (-1/2 , 9/8)
(v) Latus rectum: 4*(1/8) =1/2
vi) x intercept:The parabola y=2x^2+2x+1, does not intercept x axis as it as discriminant,(coefficient x)^2 - 4*(coefficient of x^2)*(constant term) < 0 or 2^2-4(2)(1) = -4, a negative quantity.
(vii)y intercept: Put x=0 , we get y=1. The parabola cuts y axis at the coordinates , (0,1) , i.e 1 unit above x axis.
viii) Is the origin a point on the parabola: (x,y)=(0,0) does not satisfy the equation y=2x^2+2x+1. So, the parabola is not passing through the origin.
Now , we got a good idea of the given parabola. Now we can plot the parabola by the above features or by actually giving certain values to x and obtaining the y values as below and plot it on graph sheet.
x:::::::::::::: value: -4 , -3 , -2, 1, -1/2, 0, 1 , 2, 3
2x^2+2x+1 value: 25 , 13 , 5, 1, 1/2, 1, 5 , 13, 25
From the values of x and y you can see that the parabola is symmetrical about x=-1/2 and its value is minimum 1/2 at x=-1/2.
Posted by neela on August 19, 2009 at 4:31 PM (Answer #3)
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