1. Find the sum of the constants a, h, k and such that
\(2x^2 - 8x + 7 = a(x - h)^2 + k\)
for all real numbers x.
2. Find the vertex of the graph of the equation \(x -y^2 + 6y =8\).
3. Find the area of the region enclosed by the graph of the equation \(x^2 + y^2 = 4x + 6y+13\).
1. 2x2 - 8x + 7
Factor 2 out of the first two terms.
= 2(x2 - 4x) + 7
Add and subtract (4/2)2 , which is 4 .
= 2(x2 - 4x + 4 - 4) + 7
Factor x2 - 4x + 4 as a perfect square trinomial.
= 2( (x - 2)2 - 4 ) + 7
Now distribute the 2 to the two terms in parenthesees.
= 2(x - 2)2 - 8 + 7
= 2(x - 2)2 - 1
So... a = 2 , h = 2 , k = -1 ...and... 2 + 2 + -1 = 3
2.
x - y2 + 6y = 8
Add y2 to both sides, subtract 6y from both sides.
x = y2 - 6y + 8
Complete the square on the right side.
x = y2 - 6y + 9 - 9 + 8
x = (y - 3)2 - 1
The vertex is at (-1, 3)
3.
x2 + y2 = 4x + 6y + 13 This is the equation of a circle. Let's get it in standard form...
x2 - 4x + y2 - 6y = 13 Add 4 and 9 to both sides.
x2 - 4x + 4 + y2 - 6y + 9 = 13 + 4 + 9
(x - 2)2 + (y - 3)2 = 26
Let the radius of the circle be r , now we can see that r2 = 26 .
area of the circle = pi * r2 = pi * 26 = 26pi sq. units