SOLUTION: graph the following parabola: (x-4)^2=4(y-1)

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Question 523603: graph the following parabola: (x-4)^2=4(y-1)
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
(x-4)^2 = 4(y-1)
.
x^2 -8x + 16 = 4y -4
.
4y-4 = x^2 -8x + 16
.
4y = x^2 -8x +20
.
y = (1/4)*x^2 -2x + 5
.
+graph%28500%2C500%2C-10%2C10%2C-20%2C20%2C1%2F4%2Ax%5E2-2%2Ax%2B5%29+
.
You will notice the parabola does not cross or touch the x-axis, so there are no solutions with real numbers. You can confirm this with the quadratic equation.
.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 0.25x%5E2%2B-2x%2B5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-2%29%5E2-4%2A0.25%2A5=-1.

The discriminant -1 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -1 is + or - sqrt%28+1%29+=+1.

The solution is x%5B12%5D+=+%28--2%2B-+i%2Asqrt%28+-1+%29%29%2F2%5C0.25+=++%28--2%2B-+i%2A1%29%2F2%5C0.25+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+0.25%2Ax%5E2%2B-2%2Ax%2B5+%29