SOLUTION: My problem is:Identify the completing the squares, the vertex, the y-intercept and the axis of symmetry. 1. y=x^2+6x-11 2. y=3x^2-x+4 3. y=2x^2+7x-3 4. y=2x^2-8x-12 I need

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: My problem is:Identify the completing the squares, the vertex, the y-intercept and the axis of symmetry. 1. y=x^2+6x-11 2. y=3x^2-x+4 3. y=2x^2+7x-3 4. y=2x^2-8x-12 I need      Log On


   

Question 287399: My problem is:Identify the completing the squares, the vertex, the y-intercept and the axis of symmetry.
1. y=x^2+6x-11
2. y=3x^2-x+4
3. y=2x^2+7x-3
4. y=2x^2-8x-12
I need step by step answers
my email address is gdkpw@hotmail.com
Thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The best way to think of it is:
The axis of symmetry is x+=+k where k
is midway between the roots of the equation.
Then you find the vertex by plugging in x=a
back into the equation to get y , then x%2Cy
is the vertex.
An easy formula is: if the equation is in the form
y+=+ax%5E2+%2B+bx+%2B+c, then the axis of symmetry is at
x+=+-b%2F%282a%29
The y-intercept is at y+=+c
---------------------------------------
I'll do your 1st problem:
y+=+x%5E2+%2B+6x+-+11
The y-intercept is at y+=+-11
Now find roots:
x%5E2+%2B+6x++-+11+=+0
x%5E2+%2B+6x+=+11
x%5E2+%2B+6x+%2B+%286%2F2%29%5E2+=+11+%2B+%286%2F2%29%5E2
x%5E2+%2B+6x+%2B+9+=+11+%2B+9
%28x+%2B+3%29%5E2+=+20
x+%2B+3+=+sqrt%2820%29
x+=+-3+%2B-sqrt%2820%29
The axis of symmetry is midway between the 2 roots which are:
root1 =+-3+%2B+sqrt%2820%29
root2 = -3+-+sqrt%2820%29
-3 is midway between these roots, so
x+=+-3 is the axis of symmetry
And the simple formula x+=+-b%2F%282a%29 gives the same answer:
a+=+1
b+=+6
x+=+-6%2F%282%2A1%29
x+=+-3
Now plug this result back into the equation:
y+=+x%5E2+%2B+6x+-+11
y+=+%28-3%29%5E2+%2B+6%2A%28-3%29+-+11
y+=+9+-+18+-+11
y+=+-20
So, the vertex is at (-3,-20)
I'll plot the equation:
+graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+x%5E2+%2B+6x+-+11%29+