Find the point on the graph of y=2x-4 that is closest to the point (1,3). (Optimization equation)
To find the point on the graph of y = 2x - 4 that is closest to the given point (1,3), we can use an optimization approach. The key idea is to minimize the distance between the two points.
Step 1: Determine the distance between any point (x, y) on the graph y = 2x - 4 and the given point (1,3). We can use the distance formula, which is given by:
Distance = √[(x2 - x1)^2 + (y2 - y1)^2]
Substituting the values, we get:
Distance = √[(x - 1)^2 + (2x - 4 - 3)^2]
Simplifying further,
Distance = √[(x - 1)^2 + (2x - 7)^2]
Step 2: To find the point on the graph where the distance is minimized, we need to minimize the distance function. So, we can find the minimum of the distance function by differentiating it with respect to x and setting the derivative equal to zero.
Differentiating the distance function with respect to x:
d/dx [√[(x - 1)^2 + (2x - 7)^2]] = 0
Step 3: Solve the equation obtained in Step 2 to find the value(s) of x where the distance function is minimized.
Solving the derivative equation:
0 = [√[(x - 1)^2 + (2x - 7)^2]]'
To simplify this equation, we can square both sides of the equation to eliminate the square root:
0 = [(x - 1)^2 + (2x - 7)^2]
Expanding and simplifying the equation, we get:
0 = x^2 - 3x + 2
Step 4: Solve the quadratic equation obtained in Step 3 to find the x-coordinate(s) where the distance is minimized. In this case, we can factor the quadratic equation:
0 = (x - 1)(x - 2)
Setting each factor equal to zero and solving for x, we get:
x - 1 = 0 --> x = 1
x - 2 = 0 --> x = 2
Step 5: Substitute the values of x obtained in Step 4 back into the original equation y = 2x - 4 to find the corresponding y-coordinate(s) of the point(s) on the graph.
For x = 1:
y = 2(1) - 4 --> y = 2 - 4 --> y = -2
For x = 2:
y = 2(2) - 4 --> y = 4 - 4 --> y = 0
Therefore, there are two points on the graph y = 2x - 4 that are closest to the given point (1,3): (1, -2) and (2, 0).
D^2=f(x)= (x-1)^2+((2x-4)-3)^2
f'(x)= 2(x-1)(1)+2(2x-7)(2)
0= 2x-2+(4x-14)2
0= 2x-2+8x-28
0= 10x-30
30=10x
x=3
y=2(3)-4
y=2
(3,2)
the square over D doesn't matter.