Question

f(x)=-x²+8x-13 find stationary point of a curve​

Answer 1

Step-by-step explanation:

Just taking y=x^2 (the “8x+13″ determining how the parabola will be moved and how wide it is on the graph), since it will be opening upwards there will be a minimum on the bottom, the “middle of the parabola”. The way of determining whether something is a minimum or maximum is in relation to the y-value. In other words, for a standard coordinate plane maximums are the highest spot(s), and minimums are the lowest spots.

And by considering the gradient of either side of it (using them to “prove” the spot is the minimum), let's start at a point on the left curve. Even if we are given neither an explicit function nor graph, if we examine it on a table of inputs and outputs, we will see that as the x-value increases the y-value decreases at each step.

At first, we just continue to check if the next y-value will be less (and if so means the previous won't qualify anymore as the minimum), and before reaching the center the y-values continue to decrease at each step.

However, when we reach the middle (even if we don't know yet, and expect the next y-value to decrease as well) after that we will see the y-values begin to increase, each greater than the one before; for any y=x^2, the greater the absolute value of the x then the greater the y. Now that the graph turns and goes upward, the value after the value after the minimum will of course not be less than the center. And since the gradient is positive forever, it means that regardless of how far you go for x the y-value will never decrease (and nothing will be less than the center, only greater).

Answer 2

Step-by-step explanation:

to find stationary point we should differentiate the given curve and should equal to zero

f'(x)= -2x+8 = 0

-2x = -8

x = 4

let f(x) be y .Now , if we keep x = 4 in y

y = -(4^2)+8(4)-13

= -16 +32-13

= 3

so, stationary point is (4,3)

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