Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
3. Graphing Radical Functions
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Exercise 68 Page 258

Practice makes perfect
a To complete the square, we have to add some constant to both sides of the equation so that the right-hand side becomes a perfect square trinomial. For an equation of the form y=x^2+ bx, we have to add ( b2)^2 to both sides in order for the expression on the right-hand side to be a perfect square.
y+(b/2)^2&=x^2+ bx+(b/2)^2_(perfect square) Before we can apply this to our equation, we have to divide both sides by π so that r^2 has a coefficient of 1. S=π r+π r^2 ⇔ S/π=r+r^2 Since r has a coefficient of 1, we realize that we have to add ( 12)^2 to both sides in order to complete the square.
S/Ď€=r+r^2
S/Ď€=r^2+r
S/Ď€+(1/2)^2=r^2+r+(1/2)^2
Having completed the square, we can factor the right-hand side and then solve for r.
S/Ď€+(1/2)^2=r^2+r+(1/2)^2
S/Ď€+(1/2)^2=r^2+2r(1/2)+(1/2)^2
S/Ď€+(1/2)^2=(r+1/2)^2
â–Ľ
Solve for r
(r+1/2)^2=S/Ď€+(1/2)^2
r+1/2=± sqrt(S/π+(1/2)^2)

r+1/2 > 0

r+1/2=sqrt(S/Ď€+(1/2)^2)
r=sqrt(S/Ď€+(1/2)^2)-1/2
â–Ľ
Simplify right-hand side
r=sqrt(S/Ď€+1/4)-1/2
r=sqrt(S/Ď€+Ď€/4Ď€)-1/2
r=sqrt(1/π* S+1/π* π/4)-1/2
r=sqrt(1/Ď€(S+Ď€/4))-1/2
r=sqrt(1/Ď€)sqrt(S+Ď€/4)-1/2
r=1/sqrt(Ď€)sqrt(S+Ď€/4)-1/2
b To draw a graph on your calculator, you first have to press Y= and then write the function on one of the rows. Having written the function, you can press GRAPH to draw it.
To find the radius of the right circular cone with a slant height of 1 unit and a surface area of 3Ď€4 square units, we should substitute S with the given surface area and simplify.
r=1/sqrt(Ď€)sqrt(S+Ď€/4)-1/2
r=1/sqrt(Ď€)sqrt(3Ď€/4+Ď€/4)-1/2
â–Ľ
Simplify right-hand side
r=1/sqrt(Ď€)sqrt(4Ď€/4)-1/2
r=1/sqrt(Ď€)sqrt(Ď€)-1/2
r=sqrt(Ď€)/sqrt(Ď€)-1/2
r=1-1/2
r=2/2-1/2
r=1/2
The radius of the cone is 12 units.