Consider the functions in the figure below. Find the coordinates of C in terms of b.
So there are two functions. There is the y=x^2 u-shape and then a line in the middle of it with points (0,b) and (1,1) in that order from left to right. Farther to the left of point (0,b) is point C.
I have no idea how to attempt this problem.
Apparently we're supposed to come up with the formula of a line but I am confused about what it would be due to the presence of the variable. Please help!
3 answers
For better clarification, the first part to this problem had (0,2) instead of (0,b) and the answer came out to be (-2,4). I kind of guess and checked for that question.
I will assume that point C is supposed to be on the curve y = x^2
and that point (1,1) stays constant.
label the points:
A(1,1), B(0,b) and C(c,c^2) , since y = x^2
slope of AB = (b-1)/-1 or 1 - b
slope of AC = slope of AB
= (c^2 - 1)/(c-1)
then (c^2 - 1)/(c-1) = 1-b
c^2 - 1 = c - cb - 1 + b
c^2 + cb - c - b = 0
c^2 + c(b-1) - b = 0
this is a quadratic equation, using the formula
c = (1-b ± √(b-1)^2 - 4(1)(-b) )/2
= (1-b ± √(b^2 - 2b + 1 + 4b)/2
= (1-b) ± √(b^2 + 2b + 1) )/2
= (1 - b ± √(b+1)^2)/2
= (1-b + b+1)/2 OR (1-b - b - 1)/2
= 1 or -b
if c = 1, we of course get our point A back
if c = -b,
the point becomes C(-b,b^2)
notice that your numerical example of (0,2) yielding (-2,4) fits my answer
and that point (1,1) stays constant.
label the points:
A(1,1), B(0,b) and C(c,c^2) , since y = x^2
slope of AB = (b-1)/-1 or 1 - b
slope of AC = slope of AB
= (c^2 - 1)/(c-1)
then (c^2 - 1)/(c-1) = 1-b
c^2 - 1 = c - cb - 1 + b
c^2 + cb - c - b = 0
c^2 + c(b-1) - b = 0
this is a quadratic equation, using the formula
c = (1-b ± √(b-1)^2 - 4(1)(-b) )/2
= (1-b ± √(b^2 - 2b + 1 + 4b)/2
= (1-b) ± √(b^2 + 2b + 1) )/2
= (1 - b ± √(b+1)^2)/2
= (1-b + b+1)/2 OR (1-b - b - 1)/2
= 1 or -b
if c = 1, we of course get our point A back
if c = -b,
the point becomes C(-b,b^2)
notice that your numerical example of (0,2) yielding (-2,4) fits my answer
Wow, you're a genius! Thank you so much, that worked! I didn't even think about involving the quadratic equation! Thank you again!