How do you find the minimum point of a quadratic graph?

Question

For example, what is the minimum point of y = 2xsqrd - 5x + 4? I'm guessing its 4, but I want to check. Thanks

Answer 1

Differentiating the function and setting its derivative to 0 is the fastest way however if you do not know calculus then an alternate method is to complete the square.

Firstly we pull out the factor of 2 to get y=2[x^2-5/2x+2]

Next we express the term inside the square brackets as a binomial square and a number that is needed to recover the original expression when the binomial term is expanded:

y = 2[(x-5/4)^2-(5/4)^2+2]

= 2[(x-5/4)^2+7/16]

Note that I have assumed the factor in the square brackets is a perfect square which means that it is of the from (x-b/2)^2 where b is the coefficient of the x term, in this case 5/2. However I know that it is not this perfect square so I have removed the (5/4)^2 term which the expansion of (x-5/4)^2 adds on to x^2-5/2x and then added the 2 to make the term into the original expression.

We can now see where the minimum is and what is value is. The 2 is positive, the (x-5/4)^2 term is positive since it is the square of a number and the 7/16 term is positive. Since the 7/16 term cannot be altered and the factor of 2 cannot be reduced by altering x the only term that can be is the (x-5/4)^2. The minimum will hence occur when x=5/4 and this term becomes 0 since this has removed a positive term in a sum of two positive numbers multiplied by a positive constant. The minimum value is the value of y at this minimum value of x, which is

y=2(7\16)=7\8.

The co-ordinates of the minimum are then (5/4,7/8).

What you may find more interesting is using this method for the general case y=ax^2+bx+c. I will leave this for you to do yourself however I will tell you that the result is that the minimum occurs at x=-b/2a.

Jez

Answer 2

If you know how to differentiate, that's the best way. You take the gradient function and set it equal to zero, so in this case:
4x - 5 = 0 so x = 1.25
Then put 1.25 back into the original equation, so:
y = 2x1.25^2 - 5x1.25 +4 = 0.875

so coordinates of lowest point are (1.25 , 0.875)

The lowest height is 0.875.

If you don't know how to differentiate, try completing the square.

Completing the square:

2x^2 - 5x +4 = 2(x- 5/4)^2 - 25/16 + 4

Since something squared can only be positive or zero, the lowest point will be when the 2(x - 5/4)^2 term is zero, i.e. when x = 5/4, and so you get the same result. This method doesn't require differentiation, but this can only be used on quadratics, differentiation will work for cubics or anything.

But for any quadratic in the form ax^2 + bx +c, where a, b and c are constants (in this case a=2, b= -5, c=4): the lowest point is when x = -b/2a.

Answer 3

y = 2x^2 - 5x + 4
y' = 4x - 5
y'' = 4=> The only turning point is a minimum point.

At the turning point, gradient=0 => y'=0
4x - 5 = 0
x = 5/4, so it is not 4.

If you do not know how to differentiate, you can use completing the squares:

y = 2x^2 - 5x + 4
y = 2(x^2 - 5/2 x + 2)
y = 2[(x - 5/4)^2 + 7/16]
y = 2(x - 5/4)^2 + 7/8

So the minimum point is (5/4, 7/8).

Answer 4

first take the derivative of the equation and then set it to 0.

The derivative is 4x-5

If you set 4x-5 equal to 0, then x = 5/4, this is where the point is minimum. Its only a minimum for a quadratic if the coefficient in front of the xsqrd is positive (i.e. 2 in this case). If it had been negative then it would be a maximum.

Answer 5

Using calculus-
dy/dx = 4x - 5
For a minimum, dy/dx = 0 so 4x -5 = 0 and x = 1.25
Then y = 2 x 1.25 squared - 5 x 1.25 + 4
y = 3.125 - 6.25 + 4 = 0.875

Alternatively you could draw a graph- but you will only get an approximate answer that way.

Answer 6

Using calculus, setting the first derivative equal to zero will give a minima or maxima. If the second derivative < 0 you have a maxima, if the second derivative > 0 you have a minima.

So for this equation, the first derivative,

dy/dx = 4x - 5

This equals zero when x=5/4 = 1.25 and consequently y = 0.875

The second derivative, d^2y/dx^2 = 4. A positive number. Hence the point (1.25, 0.875) is indeed a minima of the function.

Answer 7

differentiate the eq (this gives u the gradient which should be = 0 )
e.g 4x - 5 = 0
therefore x = 5/4 and to find value of y substitute value of x in the eq
y = 2xsqrd - 5x + 4

Answer 8

the minimum is also the root of the derivative (since you have a positive quadratic):
(2x^2-5x+4)' = 4x-5
the root of that is 5/4

Answer 9

differentiate it then put dy/dx=0

dy/dx=4x-5
0=4x-5
5=4x
5/4=x

then you put it on the original equation to get the y value

y=2(5/4)^2-5(5/4)+4
y=7/8

Answer 10

Graph the parabola and the lowest point on it is the minimum.