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
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)