Write an equation for a circle with a diameter that has endpoints at (–3, –2) and (7, –6). round to the nearest tenth if necessary.

The standard form for the equation of a circle is :

 (x−h)^2+(y−k)^2=r2  EQ(1)

 where handk are the x and y coordinates of the center of the circle and r is the radius.

 The center of the circle is the midpoint of the diameter.

 So the midpoint of the diameter with endpoints at (−3,-2)and(7,-6) is :

 ((−3+7)/2,(-2+(-6))/2)=(2,-4)

 So the point (2,-4) is the center of the circle.

 Now, use the distance formula to find the radius of the circle:

 r^2=(−3−(2))^2+(-2-(-4))^2=25+4=29

 ⇒r=√29

 Subtituting h=2, k=-4 and r=√29 into EQ(1) gives :

 (x-2)^2+(y+4)^2=29

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

if x = 1, then the answer is 3.5