Give the center of the circle with equation x^2+2x+y^2-10y+22=0?

Question

1.Give the center of the circle with equation x^2+2x+y^2-10y+22=0

2.Several values of the two functions f and g are listed below:
x 4 5 6 7
f(x) 7 6 5 4

x 4 5 6 7
g(x) 6 7 4 5
Find (f × g)(6)

3. An object is projected upward from the top of a tower. Its distance in feet above the ground after t seconds is given by s(t)=-16t^2+64t+80. How many seconds will it take to reach ground level?

4.Find an equation for the line with y-intercept 3 that is perpendicular to the line:
y=2/3 x - 4

5.Determine the vertex of y=x^2-8x+22.

6.Find the maximum value of y=-x^2+6x

Answer 1

5. x-coordinate of vertex is -b/2a so 8/2 or 4. Substitute in for x to find y. 16-32+22 = 6. Vertex is (4, 6).

6. Maximum value of vertex opening downward will be the vertex. Again, x-coordinate of vertex is -b/2a so -6/-2 which is 3. Substitute to get 9. Vertex is (3, 9).

Answer 2

flow the 9 to the right aspect, then finished the sq. in x and y on the left: x^2 + 2x + ( ) + y^2 + 4y + ( ) = 9 x^2 + 2x + a million + y^2 + 4y + 4 = 9 + a million + 4 (x + a million)^2 + (y + 2)^2 = 14 middle is (-a million, -2), radius is sqrt(14)

Answer 3

x^2+2x+1 +y^2-10y+25=-22+26
(x+1)^2 + (y-5)^2 = 4
center = (-1,5) and radius = 2