Consider the function=4x^2-bx+(5-a), where a,b are constant change the function in the form A(x-B)^2+C where A,B and C are constant
Anonymous
+130540 4x^2-bx+(5-a) ...I'm assuming that the "optimal value" occurs at the vertex
Taking the derivative of this, we have
8x - b .... this gives the slope at any point...and since the slope at the vertex = 0 and the x value of the vertex = 10, we have
8(10) - b = 0 → b = 80 and substituting this back into the original function, we can solve for "a" thusly:
15 = 4(10)^2 - 80(10) + (5 - a) simplify
15 = 400 - 800 + 5 - a rearrange
a = 400 - 800 + 5 - 15 = - 410
So a = -410 and b = 80
And our function is
y = 4x^2 -80x + 415
Here's the graph.....https://www.desmos.com/calculator/qhpphohunu
+130540 4x^2-bx+(5-a) ...I'm assuming that the "optimal value" occurs at the vertex
Taking the derivative of this, we have
8x - b .... this gives the slope at any point...and since the slope at the vertex = 0 and the x value of the vertex = 10, we have
8(10) - b = 0 → b = 80 and substituting this back into the original function, we can solve for "a" thusly:
15 = 4(10)^2 - 80(10) + (5 - a) simplify
15 = 400 - 800 + 5 - a rearrange
a = 400 - 800 + 5 - 15 = - 410
So a = -410 and b = 80
And our function is
y = 4x^2 -80x + 415
Here's the graph.....https://www.desmos.com/calculator/qhpphohunu
