catteyes Messages 9 Reaction score 0 Thread starter Jan 7, 2008 #1 Homework Statement y=x^2[SQ.RT.(9 - x^2)] Homework Equations The Attempt at a Solution 0= x^2 (+/- 3 - x) 0=x(x+3)(x-3) x=0 x=3 x=-3 y=0[SQ.RT.(9-0)] y=SQ.RT.(9) y=+/-3 y= 3 y=-3
Homework Statement y=x^2[SQ.RT.(9 - x^2)] Homework Equations The Attempt at a Solution 0= x^2 (+/- 3 - x) 0=x(x+3)(x-3) x=0 x=3 x=-3 y=0[SQ.RT.(9-0)] y=SQ.RT.(9) y=+/-3 y= 3 y=-3
rocomath Messages 1,752 Reaction score 1 Jan 7, 2008 #2 This question should be posted in the Precalculus subforum. Anyways, your question is find the x-int & y-int of [tex]y=x^2\sqrt{9-x^2}[/tex] correct? Your procedure for finding the y-int are correct, but your work in finding the x-int is wrong. You have [tex]\sqrt{9-x^2}[/tex] you cannot just take the square root, but your answers are right. [tex]0=x^2\sqrt{9-x^2}[/tex] [tex]x^2=0[/tex] [tex]\sqrt{9-x^2}=0[/tex] Last edited: Jan 7, 2008
This question should be posted in the Precalculus subforum. Anyways, your question is find the x-int & y-int of [tex]y=x^2\sqrt{9-x^2}[/tex] correct? Your procedure for finding the y-int are correct, but your work in finding the x-int is wrong. You have [tex]\sqrt{9-x^2}[/tex] you cannot just take the square root, but your answers are right. [tex]0=x^2\sqrt{9-x^2}[/tex] [tex]x^2=0[/tex] [tex]\sqrt{9-x^2}=0[/tex]
EnumaElish Science Advisor Messages 2,348 Reaction score 124 Jan 7, 2008 #3 The X-axis intercepts are x=0, x=3, x=-3. (Start with: 0 = x^2 sqrt[(x+3)(x-3)]. Hint: Square both sides.) The Y-axis intercept(s) are easier to solve: the are found by substituting x = 0 into the formula. Last edited: Jan 7, 2008
The X-axis intercepts are x=0, x=3, x=-3. (Start with: 0 = x^2 sqrt[(x+3)(x-3)]. Hint: Square both sides.) The Y-axis intercept(s) are easier to solve: the are found by substituting x = 0 into the formula.