how do you turn the quadratic equation 1/2x^2+X-1 into the form that i can graph it?

Question

I've been stuck on this problem for a while and I think I'm doing everything right but it is not coming out like in the back of the book. The Answer in the back of the book is 1/2(x+1)^2-3/2... I keep getting 1/2(x+1/2)^2 -9/8 . I know you have to complete the square but it's not coming out right. hope you can help me.

Answer 1

y = 1/2x^2+X-1
y = 1/2{ x^2 +2x} -1
y = 1/2{ x^2 +2x +1 -1} -1
y = 1/2{ (x^2 +2x +1) -1} -1
y = 1/2{ (x +1)^2 -1} -1
y = 1/2{ (x +1)^2} -1/2 -1
y = 1/2{ (x +1)^2 } -3/2
This is a parabola having vertex (-1,-3/2)
To find the point of intersection on x-axis put y=0 in the equation y =1/2x^2+X-1
1/2x^2+X-1 =0
x^2 +2x -2 =0
Therefore by using quadratic formula
x =0.732 and -2.73
Therefore point on x-Axis are(0.73,0) and (-2.73,0) and y intercepts (0,-1) by putting x=0

Answer 2

say y=1/2x^2+x-1
for various x get values of y and
plot the graph.
x=0,y=infinity
x=-1,y=-3/2
x=1,y=1/2
x=2,y=1/8+1=9/8
we can plot the graph using these values.
We observe the graph is discontinuous at x=0

Answer 3

1/2 x^2 + x - 1 = 1/2 (x + 2x + 1) - 1/2 - 1 = 1/2 (x+1)^2 - 3/2

Vertex at (-1, -3/2)
y-intercept at (0, -1)

For y=0 (roots):
1/2 (x+1)^2 = 3/2
(x+1)^2 = 3
x = -1 +/- sqrt(3)

Easy to graph now

Answer 4

properly your prevalent style is inaccurate... must be a(x - h)^2 + ok at the start, the a will continually be the coefficient it truly is in the front of the x^2 time period at the moment...then set up for polishing off the sq.... celebration: 5x^2 -3x + 2 => 5(x^2 - 0.6 x + ___ ) + 2 + ___ polishing off the sq. can be a 0.3^2 (undergo in ideas for the 2d clean that you do the different signal because the interior and also do not ignore the prime coefficient 5 for this reason) 5(x^2 - 0.6x + 0.9) + 2 - 4.5 => 5(x-0.3)^2 - 2.5 some thing are yours.

Answer 5

Assuming that x and X represent the same variable, it can be graphed just as it is. Go to the website below, delete the stuff in the box, type your formula there, and hit the "graph it" button. You'll get a nice parabola.

Answer 6

Not clear. Why do anything to the equation? I can "graph" your equation by simply inserting values of X into it and solving for corresponding Y values. With the (X,Y) points solved, a graph can be drawn on ordinary Cartesian coordinate graph paper.

Answer 7

a=1/2
b=1
c=-1

vertex of the parabola V(r,k)

r=-b/(2a)=-1/1=-1

k=(4ac-b*b)/(4a)=(-2-1)/2=-3/2

the arms are up because a=1/2>0

Draw the parabola y=1/2. (x+1)^2 - 3/2

Answer 8

1/2(x^2+2x-2)
let us factorise x^2+2x-2=>x^2+2x+1-3
=(x+1)^2-3
so now 1/2(x^2+2x-2)=1/2(x+1)^2-3/2