| |
You are here: Experts > Science > Mathematics > Advanced Math > Algebra II.....please help me its urgent!!!
Advanced Math - Algebra II.....please help me its urgent!!!
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
Add to this Answer Ask a Question
|
|
