Question 379999: Please help me! I am stuck and don't know if I'm doing this right and it's due tomorrow!!! I am confused when I am letting y=0 and solving for x using the quadratic formula... that is where I get stumped.
y=x^2-2x-2
I need to find the vertex and any two convenient points to sketch the graph. This is what I've done so far and I will note when I get confused.
Letting y=0
{0=-(2)+- sqrt(-2)^2-4(1)(2)all over 2}
then I end up with
{x=2+- sqrt(-4)} over 2
I do not know where to get next. Do I divide the 2 into the 2 and 4? Then that would give me -1 +- 2 right??? Then I would set them to zero and have my x points?
I have solved for y by letting x=0
y=o^2-2(0)-2
y=-2
(0,-2)
I have found the vertex by using: x=b/2(a)
x=2/2(1) y=(1)^2-2(1)-2
x=1 y= -3 giving me plots of (1,-3)
I really hope someone can help me. I know it is late but I have run out of resources. Thanks D.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
 Start with the given equation.
Notice that the quadratic  is in the form of  where  ,  , and 
Let's use the quadratic formula to solve for "x":
 Start with the quadratic formula
 Plug in , , and
 Negate  to get  .
 Square to get  .
 Multiply  to get 
 Rewrite  as 
 Add to  to get 
 Multiply and  to get .
 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)
 Break up the fraction.
 Reduce.
 or  Break up the expression.
So the solutions are or
Since these values approximate to  or  , the x-intercepts are approximately (2.73205081, 0) and (-0.732050808, 0)
You have the correct vertex and y-intercept.
If you need more help, email me at jim_thompson5910@hotmail.com
Also, feel free to check out my tutoring website
Jim
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