how can i find the local minimum

Question

P(x) = −x2 + 2x find the local minimum (x,y)

Answer 1

The graph of the function P(x) = -x² + 2x is a downward opening parabola (because the coefficient of the x² term is negative).

That means that locating the vertex will provide the location of the local max (this is also a global max for the function, i.e. the function never generates a y-value greater than the y-value of the vertex).

There are several ways to locate the vertex.

One way is to determine the axis of symmetry using the formula x = -b/2a (in this case a = -1 and b = 2).

So the axis of symmetry is x = -2/(2)(-1) = 1. Since the vertex lies on the axis of symmetry, the x-coordinate of the vertex is also 1. Plug that value into the function to determine the y-coordinate of the vertex.

P(1) = -(1)² + 2(1) = 1.

So the vertex is at (1,1). The max values occurs when x = 1 and the max value is 1 too.

P(x) can also be written as:

P(x) = -x (x - 2). This tells you the roots are at (0,0) and (2,0). The axis of symmetry passes through the midpoint of the segment joining the roots (1, 0).

Yet another way to find the vertex is to transform the function into vertex form by a process called completing the square.

P(x) = -(x² - 2x + ) [ factor out the -1 leading coefficient, leaving a space to convert the expression inside the parentheses to a perfect square trinomial -- by completing the square).

P(x) = - (x² -2x + 1) + 1 [the +1 inside the parentheses completes the square. By placing that +1 inside the parentheses a -1 has really be added because of the -1 outside, so add a +1 outside the parentheses to keep the equation in balance.

Finally rewrite the perfect square trinomial as a binomial squared:

P(x) = - (x - 1)² + 1

This is called vertex form because it is easy to identify the vertex (1,1). Note that when x = 1 the the first term has a value of zero, so the function value is 0 + 1 = 1. If x is any other value then the first term has a negative value so the result will be a y-value less than 1, i.e. the max value of the function is the y-coordinate of the vertex.

desmos.com/calculator/kith1niagf

Answer 2

P(x) = -x^2+2x

take the derivative and set equal to zero

y' = -2x+2 = 0

-x +1 = 0

x =1

y = -(1)^2 +2(1) =-1+2 = 1

the maximum, local and absolute, point is (x,y) = (1,1)

there is no local minimum or absolute minimum

as the graph is a downward opening parabola with vertex = max point = (1,1)

end behavior is to negative infinity

you might try to say the minimum is negative infinity, but infinity is not a real number

the P(x) value only approaches infinity and never reaches it, and it's not a local minimum in any sense of the word

alternative method is convert the quadratic into vertex form by completing the square

y= -x^2 + 2x

= -(x^2 +2x + 1) + 1

=-(x+1)^2 + 1

vertex is (1,1) = the maximum and = the only extremum, as there is no minimum of any kind

y = a(x-h)^2 + k is the vertex form, where (h,k) is the vertex. in this problem (h,k) = (1,1)

whenever y=ax^2+bx +c has a<0, there is a maximum, but no minimum