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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Camo on March 20, 2011, 07:36:21 pm

Title: Camo'd question thread ;D
Post by: Camo on March 20, 2011, 07:36:21 pm: :)

So a few questions.

will the following touch the x-axis?
a)f(X)=x^2-5x+2
b)f(X)=3x^2+2x+5
Title: Re: Camo'd question thread ;D
Post by: Water on March 20, 2011, 07:39:35 pm: a)f(X)=x^2-5x+2
b)f(X)=3x^2+2x+5

Find the determinant of each quadratics. If it has one solution, it is touching the x axis.
Title: Re: Camo'd question thread ;D
Post by: pi on March 20, 2011, 07:41:32 pm: +1 to Water's help, except its actually 'discriminant' :) (as ClimbTooHigh pointed out)

$Discriminant = \Delta = b^2 -4ac \ \ where \ \ y=ax^2 + bx + c$
Title: Re: Camo'd question thread ;D
Post by: BubbleWrapMan on March 20, 2011, 07:47:56 pm: Discriminant.
Title: Re: Camo'd question thread ;D
Post by: pi on March 20, 2011, 07:49:30 pm: Quote from: ClimbTooHigh on March 20, 2011, 07:47:56 pm
Discriminant.

True, DAMN THOSE MATRICES!
Title: Re: Camo'd question thread ;D
Post by: Camo on March 20, 2011, 07:55:12 pm: sketch f(X)=2(x+1)^2 this will have intercepts at (-1,0) and (0,5)?
Title: Re: Camo'd question thread ;D
Post by: Camo on March 20, 2011, 07:55:47 pm: sorry (0,4).
Title: Re: Camo'd question thread ;D
Post by: pi on March 20, 2011, 07:56:47 pm: Quote from: Camo on March 20, 2011, 07:55:12 pm
sketch f(X)=2(x+1)^2 this will have intercepts at (-1,0) and (0,5)?

(-1,0) and (0,2)

y-int when x=0
f(0)= 2(1)^2 = 2
Title: Re: Camo'd question thread ;D
Post by: Camo on March 20, 2011, 07:59:30 pm: Rohitpi yours answers are correct for f(x)=2(x-1)^2 i missed up.
Title: Re: Camo'd question thread ;D
Post by: Greatness on March 20, 2011, 08:12:22 pm: f(x)=2(x-1)^2
X ints: (y=0) 2(x-1)^2=0 -> x-1=0 -> x=1 (1,0)(so it touches the x axis)
Y int: (x=0) f(0)=2(0-1)^2 -> y=2 (0,2)
Title: Re: Camo'd question thread ;D
Post by: Camo on March 20, 2011, 08:15:26 pm: express f(X)=x^2+3x-2 and f(X)=4x^2+8x-7 in the form y=a(x+h)^2 and hence the max or min and the range in each.
Title: Re: Camo'd question thread ;D
Post by: jasoN- on March 20, 2011, 08:20:34 pm: complete the square:

ill do for 4x^2 + 8x - 7:
y=4(x^2+2x-7/4)
y=4(x^2+2x+1^2-1^2-7/4)
y=4[(x+1)^2-1-7/4]
y=4[(x+1)^2-11/4]
y=4(x+1)^2-11
a is positive, so its a minimum (-1,-11)
range = (-11,infinity)
Title: Re: Camo'd question thread ;D
Post by: dooodyo on March 20, 2011, 10:11:59 pm: how would you find x-intercept of -2log e(x+2) ?
Title: Re: Camo'd question thread ;D
Post by: brightsky on March 20, 2011, 10:12:58 pm: Let y = -2loge(x+2)
For x-intercept, let y = 0
-2loge(x+2) = 0
loge(x+2) = 0
x+2 = e^0 = 1
x = -1
Title: Re: Camo'd question thread ;D
Post by: Camo on March 21, 2011, 11:29:41 am: the graph of y=a(x+h)^3+k has a point of zero gradient at (0,4) and passes through (1,1) find the values of a,h and k.
i know k = 4 due to the y-intercept of the zero gradient and its cubic.
Title: Re: Camo'd question thread ;D
Post by: Mao on March 21, 2011, 11:55:50 am: By the same logic, h is 0 (it's very similar to 'turning point form' for quadratics)

which leaves you y=ax^3 + 4, which need to pass through (1,1), finding a is a matter of substitution.
Title: Re: Camo'd question thread ;D
Post by: xZero on March 21, 2011, 11:58:10 am: This is the tp form so h is 0 and k is 4. Sub in x=1 and y =1 to work out a