Find the vertx, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function. f(x)=2x^2+2x+1

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kjcdb8er's profile pic

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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.

neela's profile pic

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The second degree equation of the form y= ax^2+bx+c represents a parabola.

y=2x^2+2x+1

y=2(x+1/2)^2-2(1/2)^2+1

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:

Y/2-1/2=(x+1/2)^2 or

(x+1/2)^2 = y/2-1/2= (1/2)(y-1)

(x+1/2)^2=4(1/8)(y-1) (2)

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.

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