Can someone explain how to find the vertex of a parabola using the derivative?

Question

I'm not yet in calculus, but differentiating doesn't seem too hard (I can get down the very basics)...I was just wondering if someone would be able to explain how to find the vertex of a parabola using the derivative...I really hate using complete the square...it's time consuming and boring! Using another method would also be useful to double check my answer. Also, are there any other calculus tricks that are useful for simple applications (ex. simple functions, trig, etc.). Thank you in advance.

Answer 1

y = x^2 + 1
y' = 2x
0 = 2x

<-------- - ------- 0 ------- + ---------->

Given an equation, say x^2 + 1, find the derivative using the power rule. Bring the power down in front as the coefficient, and reduce the power to 1. The derivative of a constant is always 0. After you find the derivative, set it equal to 0 and do a sign graph (you should know how to do this if you're in pre-calc). y' = 2x = 0, thus y' = 0 when x = 0. Plug in numbers less than zero and greater than zero to find the sign.

- to + = a min
+ to - = a max

Thus there is a min at x = 0. Plug in x = 0 into the equation y = x^2 + 1, and you find y = 1. So the vertex is at (0,1). This reasoning will work for any quadratic.

Answer 2

Vertex Of A Parabola

Answer 3

It is quicker and as a check to use the fact that the x coordinate of the vertex of y = ax^2+bx+c is the value
x = -b/2a. The derivative of the above is y'=2ax+b and since the derivative gives the slope and the slope of the tangent at the vertex is 0, setting the derivative = 0 and solving for x gives the formula for the vertex above.

Answer 4

Hi,
It's better demonstrated with a formula. Let's take this parabola:
y=x²-4x +3
y' = 2x -4...(this is the derivative using the rule nu^(n-1).)
Set that equal to zero and solve for x:
2x-4=0
2x=4
x=2
Now, for the y-coordinate, substitute that value into the function and solve for y:
y=(2)²-4(2)+3
=4-8+3
=-1
There's another method just as simple that does not depend on the derivative:
x=-b/(2a)...(From the form ax²+bx +c.)
For the above equation:
x=-(-4)/(2*1)
=2
y=(2)²-4(2)+3
=4-8+3
=-1
Same answer, no more work.

FE

Answer 5

The gradient of the vertex of a parabola is always zero, hence dy/dx=0.

E.g. y = ax^2 + bx + c
dy/dx = 2ax + b

Hence to find the vertex, dy/dx=0:
2ax + b = 0
x = -b/(2a)