How do you find the turning point of these equations?

👨🏻‍🏫 學術台數學

[匿名]

2009-01-12 12:04 am

y=2x^2-4x
and
y=-2x^2-3x

i forgot everything over the holidays and now I'm going back to school soon so I gotta start doing my holiday homework. All I remember was use completing the square or something.

can you show working out too?

更新1:

also, is it easier to find the x intercepts using the null factor law or the quadratic formula in this case?

回答 (9)

[匿名]

2009-01-12 12:08 am

✔ 最佳答案

You need to get the equations in the form a(x - h)² + k, where h, k is the turning point. That form can be obtained by completing the square.

y = 2x² - 4x

2(x² - 2x) {Factorise 2 out}

2 [(x - 1)² - 1] {Complete the square}

2(x - 1)² - 2 {Distribute the 2}

In this form, we can read the turning point as 1, -2

Same thing applies for your second equation?

y = -2x² - 3x

-2(x² - 1.5) {Factor -2 out}

-2[(x - 0.75)² - 0.5625] {Complete the square}

-2(x - 0.75)² + 1.125 {Distribute 2 out}

We can now read the turning point off. It is 0.75, 1.125.

And it is easier to solve these quadratics using the null factor law. ie

y = 2x² - 4x
y = 2x ( x - 2 )
Let y = 0 to find x intercepts.
2x ( x - 2 ) = 0

x must equal 0 or 2.

Nice and simple.

Hope that helps.

=D

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[匿名]

2009-01-12 12:09 am

Stationary point...

Differentiate each equation , then equate to 0 to find values of x.

Then find the 2nd differential to find the nature of the turning point by putting your values of x into the 2nd differential equation.

The value of x which give a negative value is a max and vice versa.

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[匿名]

2009-01-12 12:09 am

Differentiate the two equations - then set them equal to 0 (the turning point is when the gradient is 0 and the differential of an equation is its gradient). You should be able to work them out for yourself

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DWRead

2009-01-12 12:08 am

y = 2xÂ² - 4x
y' = 4x - 4 = 0
x = 1
(1, -2)

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[匿名]

2009-01-12 12:11 am

I'll do the first example for you.
You diffrentiate them first with respect to x to get
dy/dx = 4x-4
d^2y/dx^2 = 4

4 = turning point.

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[匿名]

2009-01-12 4:03 am

to find the turning point (or stationary point) you need to take the first derivative and equate to zero for the value for x, then use it to find y

y = 2x^2 - 4x

y' = 4x - 4
= 4(x-1) = 0

=> x =1, y = -2

therefore we have a stationary point at (1,-2)

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An ESL Learner

2009-01-12 2:58 am

y = 2x^2 - 4x (solve by using substitution)
y = -2x^2 - 3x

y = 2x^2 - 4x
-2x^2 - 3x = 2x^2 - 4x
2x^2 + 2x^2 + 3x - 4x = 0
4x^2 - x = 0
x(4x - 1) = 0

x = 0

4x - 1 = 0
4x = 1
x = 1/4 (0.25)

y = 2x^2 - 4x
y = 2(0)^2 - 4(0)
y = 2(0) - 0
y = 0

y = 2x^2 - 4x
y = 2(1/4)^2 - 4(1/4)
y = 2(1/16) - 4/4
y = 2/16 - 1
y = 1/8 - 8/8
y = -7/8 (-0.875)

â´ [x = 0 , y = 0] , [x = 1/4 (0.25) , y = -7/8 (-0.75)]

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[匿名]

2009-01-12 1:42 am

y= 2x^2 - 4x

Note: It is easier to factorise using the null factor law because when using the quadratic formula, the equation needs to be in the form :
ax^2 + bx +c

Factorise bytaking out a common factor

y= 2x ( x-2)

2x=0 or x-2=0

x=0 or x=2

The x-intercepts are 0 and 2

Turning point:

x= 0+2/2
=1
y= Substitute the x value (1) into the equation....
=2(1)^2 - 4(1)
=-2

The turning point is (1,-2)

Also, hope you know how to find the y-intercept right?
y= 2(0)^2 - 4(0)
=0

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[匿名]

2009-01-12 12:13 am

Since these are quadratic functions, you can find the axis of symmetry (which is given by x = -b/2a) and then find out the y value.

1) y = 2xÂ² - 4x
Axis of sym:
x = -b/2a
= -(-4)/2(2)
= 4/4 = 1

x = 1 ---> y = 2(1)Â² - 4(1)
= 2 - 4 = -2

Therefore there's a turning point at (1,-2)

2) y = -2xÂ² - 3x
Axis of sym:
x = -b/2a
= -(-3)/[2(-2)]
= -3/4

x = -3/4 ---> y = -2(-3/4)Â² - 3(-3/4)
= -9/8 + 9/4
= 9/8

Therefore there's a turning point at (-3/4, 9/8)

Another method you can use it differentiation.

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