SOLUTION: So this is where I am at with #s 1 and 2. Not sure if they were done correctly. The problem is that I have no idea how to graph a parabola or whatever. Please help me ANYONE!!I

Question 500591: So this is where I am at with #s 1 and 2. Not sure if they were done correctly. The problem is that I have no idea how to graph a parabola or whatever. Please help me ANYONE!!I sure would appreciate it! Thanks.....SuzieFor questions 1-2, apply the quadratic formula to find the roots of the given function, and then graph the function.
1. (6 points) f(x) = x2 + 4
F(x) = x^2 + 4
Ax^2 + bx + c = 0
X =-b+√b^2-4*a*c
2*a
X^2 + 4 = 0
Since there is no x term the 2nd coefficient is 0
X^2 + 0*x+4=0
A=1 b=0 c=4
X=0+√(0)^2-4*1*4
2*1
X =0+√0-4*1*4
2*1
X=0+√0+(-16)
2*1
X=0+√-16
2*1
X=0+4*i
2*1
X=0+4*1
2
X=0+2*I or x=0-2*i
See Graph Attached
We can see that there are no real roots

2. (6 points) g(x) = x2 + x + 12
G(x) = x^2 + x + 12
Ax^2 + bx + c = 0
X = -b+√b^2-4*a*c
2*a
X^2 + x + 12 = 0
A=1 b = 1 c = 12
X = -1+√(1)^2-4*1*12
2*1
X = -1+√1-4*1*12
2*1
X = -1+√1+(-48)
2*1
X = -1 +√-47
2*1
X = -1+i*√47
2*1
X = -1+i*√47
2
X = -1-i*√47
2
2,18 -2,14 4,32 -4,24 Vertex is 0,12

Y = x^2 +x +12 SEE GRAPH ATTACHED we can see there are no real roots
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
So this is where I am at with #s 1 and 2. Not sure if they were done correctly. The problem is that I have no idea how to graph a parabola or whatever. Please help me ANYONE!!I sure would appreciate it! Thanks.....SuzieFor questions 1-2, apply the quadratic formula to find the roots of the given function, and then graph the function.
1. f(x) = x2 + 4
2. g(x) = x2 + x + 12
..
The two equations listed above are for parabolas of the standard form: A(x-h)^2+k, with (h,k) being the (x,y) coordinates of the vertex. If the lead coefficient is positive, the parabola opens upwards, that is, curve has a minimum. Conversely, is the lead coefficient is negative, parabola opens downward and curve has a maximum.
..
For equation 1, h=0 and k=4, so you have one point (0,4), the vertex, to plot. At this point you can see there will be no real roots as the lowest point on the curve is 4. To find two more points to plot, plug in a small number like 1 in the equation:
x^2+4=1+4=5, to get a point at (1,5)
Since the curve is symmetrical about its line of symmetry, x=0 in this case, you get another point on the other side of the line at (-1,5)
You now have the vertex opening upwards and two points on either side of the line of symmetry to plot the curve for the parabola.
..
For equation 2, you must complete the square first to find the plot points.
y=x^2+x+12
y=(x^2+x+1/4)+12-1/4
y=(x+1/2)^2+47/4
Vertex: (-1/2, 47/4)
Parabola opens upwards, curve has a minimum at 47/4, therefore, no real roots.
To find more points to plot follow the method above. The numbers are more cumbersome to work with but you should have a visual of what the curve looks like.
See the two graphs below: (Red for eq. 1, and green for eq. 2}
..

..
Comment: You can always plot points one at a time to get a graph, but I believe you are past this and are learning a more efficient method for graphing parabolas and other conics. The algebra you did is ok and necessary at times and it showed the function had no real roots, but having a visual for the function always makes algebra more understandable and interesting.
RELATED QUESTIONS
I am supposed to simplify this problem and I am not sure I have done it correctly.... (answered by edjones)

Hi- I have been working on this problem and I think I am doing it right. I am about... (answered by jim_thompson5910)

Hello I am currently trying to figure out this problem. Find all real and imaginary... (answered by jim_thompson5910)

Hey there, I�m not sure if this is the correct category for this question, but I�m not (answered by ikleyn)

I'm not sure if this is the right subject line but I am working with decimal points and... (answered by stanbon)

I am working on dividing polynomials and I am stuck on this problem. I am not sure if I... (answered by ankor@dixie-net.com)

i have previously submitted this question but have not received an answer. I have had a... (answered by praseenakos@yahoo.com,jim_thompson5910)

Not sure if it fits in the right category. Here is the question. if 8 lumberjacks can... (answered by acerX)

iam not sure if this is linear and graphs or whatever! because my worksheet is mixed so i (answered by stanbon)