The line y = mx+b with b≥2 is tangent to the graph f(x) = -2(x-2)² + 2 at a point in the first quadrant. What are all possible values of b?
Well, well, well! Look who's tangling with some lines and graphs in the first quadrant! So, we want to find all possible values of b, huh? Alrighty then, let's put on our mathy hats and get to work!
Now, if the line y = mx + b is tangent to the graph f(x) = -2(x-2)² + 2, it means that these two buddies only touch each other at one lone point. So, what does that tell us? It means the slope of the line and the slope of the graph should be the same at this special point. One big happy family!
The slope of f(x) = -2(x-2)² + 2 is -4(x-2), and the slope of y = mx + b is, quite obviously, m. So, if these two slopes are equal, we have -4(x-2) = m.
Now, let's also consider the fact that the point of tangency should lie in the first quadrant. That means the x-coordinate of the point should be positive, and also the y-coordinate. Simple enough, huh?
Now, let's replace y in the line equation with the value from the graph to get rid of one variable. So, mx + b = -2(x-2)² + 2.
Here comes the fun part! Let's solve this equation for x. Expand, simplify, do your math dance, and we'll find the x-coordinate of the point of tangency. Remember, it has to be positive!
Once we have the x-coordinate, we can plug it back into the original equation -4(x-2) = m to find the corresponding slope, which, believe it or not, is m.
Done, done, and done! We now have the slope m and the x-coordinate of the point of tangency. And since we want b≥2, we can simply plug these bad boys into the equation y = mx + b, choose b to be greater than or equal to 2, and voila! You've got all possible values of b!
So, get ready to juggle those numbers, my friend, and find all the b's that make this tangential relationship tickety-boo in the first quadrant! Good luck!
To find the possible values of b, we need to find the point of tangent on the given function.
The equation of the line is y = mx + b.
The given function is f(x) = -2(x-2)² + 2.
To find the point of tangency, we need to find the values of x and y where the line and the function are equal. Hence, we can set the equations equal to each other:
mx + b = -2(x-2)² + 2
Now, let's simplify the equation and solve for x and y:
mx + b = -2(x² - 4x + 4) + 2
mx + b = -2x² + 8x - 4 + 2
mx + b = -2x² + 8x - 2
Rearranging the terms, we get:
2x² - (m-8)x + (b-2) = 0
For a tangent line, the discriminant (b² - 4ac) should be zero. Therefore, we have:
(m-8)² - 4(2)(b-2) = 0
m² - 16m + 64 - 8b + 16 = 0
m² - 16m - 8b + 80 = 0
Now, we solve this equation for m in terms of b:
m² - 16m = 8b - 80
m² - 16m + 64 = 8b - 80 + 64
(m-8)² = 8b - 16
m = 8 ± √(8b - 16)
Since the line is tangent to the graph, there will be only one possible value of m. Therefore, the discriminant (8b - 16) must be zero:
8b - 16 = 0
8b = 16
b = 2
Hence, the only possible value of b is 2.
To determine the possible values of b such that the line y = mx + b is tangent to the graph f(x) = -2(x-2)² + 2 in the first quadrant, we need to find the point of tangency between the line and the graph.
The slope-intercept form of a line is given by y = mx + b, where m represents the slope and b represents the y-intercept.
The graph of f(x) = -2(x-2)² + 2 represents a downward-opening parabola with vertex (2, 2).
To find the point of tangency, we need to equate the equations of the line and the parabola and solve the resulting system of equations.
Setting y = mx + b and f(x) = -2(x-2)² + 2 equal to each other gives us:
mx + b = -2(x-2)² + 2
Next, we can differentiate the equation of the parabola, f(x), with respect to x to find the derivative, f'(x), which represents the slope of the tangent line at any given point:
f'(x) = -4(x-2)
Since the line y = mx + b is tangent to the graph, its slope, m, must be equal to the derivative of the function at the point of tangency. Therefore, we have:
m = f'(x) = -4(x-2)
Substituting this value of m into the equation mx + b = -2(x-2)² + 2, we get:
-4(x-2)x + b = -2(x-2)² + 2
Simplifying further:
-4x² + 8x + b = -2(x² - 4x + 4) + 2
-4x² + 8x + b = -2x² + 8x - 8 + 2
-4x² + 8x + b = -2x² + 8x - 6
Now, rearrange the equation to get it in standard quadratic form:
2x² + (-8x - 4x) + (6 - b) = 0
2x² - 12x + (6 - b) = 0
For the line to be tangent to the parabola, this quadratic equation must have a discriminant of zero. Hence:
(-12)² - 4(2)(6 - b) = 0
144 - 48(6 - b) = 0
144 - 288 + 48b = 0
-144 + 48b = 0
48b = 144
b = 144 / 48
b = 3
Therefore, the only possible value of b is 3.