How to solve worded questions involving gallery of graphs?

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Can someone please explain to me how to do question 6 and 7?
Thank you very much!

1 Answer
kubik98
Jan 2, 2018

See explanation

Explanation:

6.a:

#f: y=sqrt(x-b)+c#
#l: y=x#

If this curve and line must meet then they must equal to each other:
#x=sqrt(x-b)+c#

#(x-c)^2=(sqrt(x-b))^2#

#x^2-2xc+c^2=x-b#

#x^2-2xc-x+c^2+b=0#

Taking out minus and x.

#x^2-(2c+1)x+c^2+b=0#

Since it's the point #(a,a)# then #x=a#. We substitute and then it satisfies given equation:
#a^2-a(2c+1)+c^2+b=0#

6.b.1:

#y=sqrt(x-b)+c#
#y=x# is a tangent in the point which satisfies this equation of curve: #color(blue)1x^2color(green)(-(2c+1))x+color(red)(c^2+b)=0#

To calculate zero points of quadratic funtion, we use: #x_(1,2)=(-b+-sqrtD)/(2a)#

where: #D=b^2-4ac#
if:
#D>0quad=># function has 2 zero points
#D=0quad=># function has only 1 zero points
#D<0quad=># function has none zero points

Tangent line crosses curve only once, so we use #D=0#
Substitute:
#0=color(green)((2c+1))^2-4color(red)((c^2+b))#

#0=cancel(4c^2)+4c+1cancel(-4c^2)-4b#

#4b-1=4c#

#c=(4b-1)/4#

6.b.2:

#T[x_0,y_0]# is a point where a line #y=x# touches curve #y=sqrtx-1/4# as a tangent.

We use the same logic as above:
#x=sqrtx-1/4#

#(x+1/4)^2=x#

#x^2+(2x)/4+(1/4)^2=x#

#x^2+x1/2-x+1/16=0#

#x^2-x1/2+1/16=0#

#x_0=(1/2+-sqrt(1/4-4*1/16))/(2)#

#x_0=(1/2+-sqrt(0))/(2)#
(Note that D=0, which means we have only one possible sollution=>only one tangent exists)

#x_0=1/4#

#y_0=x_0=># a point which we look for is: #T[1/4,1/4]#
graph{sqrtx-1/4 [-10, 10, -5, 5]}

6.c:

Let me skip that. It's just applying this:
#D>0quad=># function has 2 zero points (for first point)
#D=0quad=># function has only 1 zero points (for second point)
#D<0quad=># function has none zero points (for third point)

7.b:

If the line must meet the curve once we use exactly the same process as above:

#y=kx#
#y=sqrtx-1#
#k=?#

#kx=sqrtx-1#

#(kx+1)^2=x#

#k^2x^2+2kx+1-x=0#

#k^2x^2+x(2k-1)+1=0#

#0=(2k-1)^2-4k^2#

#0=cancel(4k^2)-4k+1-cancel(4k^2)#

#k=1/4#

7.a:

We have the same equation: #k^2x^2+x(2k-1)+1=0#
Now we have to find 2 possible meeting points. #D>0#

#0<-4k+1# (we have calculated that, see above)

#k<1/4#
but that is not he end. k must have some minimum value. squareroot is always increasing therefore that line must be increasing too and that is true only when k>0. So the answer is an interval:
#k in (0,1/4)#

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