SOLUTION: please help. question: consider the parabola ( i know that on a graph thats the if you say smiley face on a graph) with the equation y+5=2(x+2)^2. a. What is the vertex? (i t

Algebra ->  Functions -> SOLUTION: please help. question: consider the parabola ( i know that on a graph thats the if you say smiley face on a graph) with the equation y+5=2(x+2)^2. a. What is the vertex? (i t      Log On


   



Question 634327: please help.
question: consider the parabola ( i know that on a graph thats the if you say smiley face on a graph) with the equation y+5=2(x+2)^2.
a. What is the vertex?
(i thought the vertex was the (x+2) but im completely not sure)
b. what are its x-intercepts?
(i was lost after that to find the x-intercept wouldnt i have to take everything from the right expect the x to find the intercept) please help me their is a third question but i know it says no graphs so i will not include c.)

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
y+5=2(x+2)^2 (y = 0 ⇒ (x+2)^2 = 5/2 and x-intercepts are: -2 ± sqrt%285%2F2%29
y =2(x+2)^2 - 5 , V(-2,-5)
the vertex form of a Parabola opening up(a>0) or down(a<0)is: y=a%28x-h%29%5E2+%2Bk
where is the vertex and x = h is the Line of Symmetry