Hi,  

the vertex form of a parabola,  where(h,k) is the vertex

vertex (-1,-108)

y =a(x+1)^2 -108

using y-intercept pt(0,-105) to find a

-105 = a(1)^2 -108

3 = a

y = 3(x+1)^2 -108  

finding x-intercepts when y = 0

0 = 3(x+1)^2 -108 

108 = 3(x+1)^2

 36 = (x+1)^2

  6 = (x+1)  x = 5

 -6 = (x+1)  x -7

 x-intercepts are Pt(5,0) and Pt(-7,0) 

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
 Find the X-intercept of the parabola with vertex (-1,-108) and y-intercept of (0,-105).

Use the form for the parabola:

y = a(x - h)² + k with vertex (h,k) = (-1,-108)

y = a[x - (-1)]² + (-108) 
y = a(x + 1)² - 108

Since y-intercept (0,-105) must satisfy the equation
we substitute 0 for x any -105 for y:

   y = a(x + 1)² - 108
-105 = a(0 + 1)² - 108
-105 = a(1)² - 108
-105 = a*1 - 108
-105 = a - 108
   3 = a

So we substitute 3 for a in

y = a(x + 1)² - 108

and get

y = 3(x + 1)² - 108

That's the equation of the parabola

To find the x intercept we substitute 0 for y:

  0 = 3(x + 1)² - 108
108 = 3(x + 1)²

Divide both side by 3

 36 = (x + 1)²

Use the principle of square roots:
  __
36 = x + 1

  ±6 = x + 1

Using the +6

   6 = x + 1
   5 = x

Using the -6

  -6 = x + 1
  -7 = x

So there are two x-intercepts (5,0) and (-7,0)



Edwin
 

 

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