SOLUTION: Please find the -y intercept x intercepts if any vertex and line of symetry of the following quadratics.

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Question 145458: Please find the -y intercept
x intercepts if any
vertex
and line of symetry of the following quadratics.


1.y=x%5E2%2B2x%2B12

2.y=x%5E2-6x%2B5

3.y=x%5E2-8x%2B15

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The y intercept is at x = 0. f(0) is:
1. 12
2. 5
3. 15
1. y+=+x%5E2%2B2x%2B12+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B12+=+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=%282%29%5E2-4%2A1%2A12=-44.

The discriminant -44 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 -44 is + or - sqrt%28+44%29+=+6.6332495807108.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B12+%29

2. x%5E2-6x%2B5=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-6x%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=%28-6%29%5E2-4%2A1%2A5=16.

Discriminant d=16 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--6%2B-sqrt%28+16+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-6%29%2Bsqrt%28+16+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-6%29-sqrt%28+16+%29%29%2F2%5C1+=+1

Quadratic expression 1x%5E2%2B-6x%2B5 can be factored:
1x%5E2%2B-6x%2B5+=+%28x-5%29%2A%28x-1%29
Again, the answer is: 5, 1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-6%2Ax%2B5+%29

3. x%5E2-8x%2B15=0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B15+=+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=%28-8%29%5E2-4%2A1%2A15=4.

Discriminant d=4 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--8%2B-sqrt%28+4+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-8%29%2Bsqrt%28+4+%29%29%2F2%5C1+=+5
x%5B2%5D+=+%28-%28-8%29-sqrt%28+4+%29%29%2F2%5C1+=+3

Quadratic expression 1x%5E2%2B-8x%2B15 can be factored:
1x%5E2%2B-8x%2B15+=+%28x-5%29%2A%28x-3%29
Again, the answer is: 5, 3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B15+%29