Expert: Ahmed Salami - 11/7/2009

Question
21. Find values of a and k for a parabola containing the points given:  
y - k = a(x + 3)^2;  (-5,1) and (1,7)

20. If the parabola y + 5 = a(x + 2)^2 has a y-intercept 4, find a.

18. Find an equation y - k = a(x - h)^2 for the parabola described:  
vertex:  (-3,4);  x-intercept:  -1

17. Find an equation y - k = a(x - h)^2 for the parabola described:
vertex:  (-2,6);  y-intercept:  -2  

Answer
Hi Krystal,
21) y - k = a(x + 3)²
At (-5,1)
1 - k = a(-5 + 3)²
     = a(-2)²
     = 4a
k = 1 - 4a
At (1,7)
7 - k = a(1 + 3)²
     = a(4)²
     = 16a
k = 7 - 16a
Therefore,
1 - 4a = 7 - 16a
12a = 6
a = 1/2
k = 1 - 4a
 = 1 - 2
 = -1

20) y + 5 = a(x + 2)²
A y-intercept of 4 means that y = 4 when x = 0, so
4 + 5 = a(0 + 2)²
9 = a(2)²
9 = 4a
a = 9/4

18) y - k = a(x - h)²
At the vertex, x = h
So, h = -3 and
y - k = a(h - h)²
y - k = 0
y = k
k = 4
At the x-intercept,
x = -1 and y = 0
Therefore,
0 - 4 = a(-1 - -3)²
-4 = a(-1 + 3)²
-4 = a(2)²
-4 = 4a
a = -1
The equation is
y - 4 = -(x + 3)²
y = 4 - (x + 3)²

17) y - k = a(x - h)²
At the vertex, x = h
So, h = -2 and
y - k = a(h - h)²
y - k = 0
y = k
k = 6
At the y-intercept,
x = 0 and y = -2
Therefore,
-2 - 6 = a(0 - -2)²
-8 = a(0 + 2)²
-8 = a(2)²
-8 = 4a
a = -2
The equation is
y - 6 = -2(x + 2)²
y = 6 - 2(x + 2)²

Note that for y - k = a(x - h)²
y = a(x - h)² + k
(x - h)² is always positive and is minimum when it is zero, i.e
x - h = 0
x = h
which explains the condition at the vertex.

Regards