Easiest way to find the minimum point?

Question

I'm trying to figure out the easiest way to find the Minimum point of:

 

f(x)=2x^2+20x+12

Answer 1

The EASIEST way is to recognize that the graph is that of a parabola, and it opens upwardly, so its minimum is at its vertex.  The formula for x coordinate of vertex is x_vertex = -b/(2a)  for  y = ax²+bx+c.  The minimum occurs at that computed vertex's x, and the minimum is the y computed for that x.

Answer 2

CPM math also teaches students to change the quadratic from 

Standard form to Graphing form:  y = a( x - h ) + k

 

vertex is (h, k), 'a' tells the narrowness or wideness of the graph

 

Changing the Quadratic:

 

y = 2x² + 20x +12    Complete the square using the x-terms

 

y = 2 (x² + 10x) + 12

 

y = 2 (x -5 )²  yields a constant term of 50 ≠ 12, so use -38 for constant

 

Graphing form for all Functions is taught in CPM math, this makes it far

easier for students to know what type of curve the equation will be and 

to graph it quite easilly!

 

y = 2( x-5)² -38   This is the graphing form of the same quadratic,

 

From this equation, student can see the vertex is (5, -38)

Answer 3

Hello Gates,

Good time to post a question!

Now Mark, a great tutor, gave an approach using algebra.

and Bam, another great tutor, gave an approach using calculus.

 

David F

Answer 4

In f(x) = ax² + bx + c the mas/min occurs at -b/2a

In the function given, -20/(2), -20/4, -5

The minimum occurs at x = -5

Now you can substitute and determine the minimum value.

Answer 5

The easiest way if to use the first derivative. f'(x) = 0 at the minimum. Therefore, 4x + 20 = 0 ⇒ x = -5 and the image f(x) = 50 - 100 + 12 = -38 ;

The minimum point is located at (-5 ; -38) on the Cartesian plane.