For a linear function, you want it in y= mx + b form
They give you the slope your x, and when they say f(7)=1 that's like saying when x=7, y=1, or the ordered pair (7,1). To make it in y=mx+b, you can plug these coordinates in to solve for b. So, 1=7(7)+b, after you plug in slope and your point. Do math, 1=49+b, b=-48
That's how I think it's done
y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
Answer:
it would move 4 in the negative direction
(-1,1) (6,9)
distance = root (x1-x2)²+(y1-y2)²
so root (-1-6)²+(1-9)²= root 113 or 10.6 ≈11
the answer is B. 11.18 units
