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#### karush

##### Well-known member

- Jan 31, 2012

- 2,685

$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,685

$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's

- Mar 1, 2012

- 658

$$y = x^2+1 \textit{and } y = x- x^2$$

Ok I think what question is ... The vertical distance between vertex's

I interpret the question as the minimum of $|y_1 -y_2|$ where $y_1=x^2+1$ and $y_2=x-x^2$

... which would be a distance equal to 7/8

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- Jan 31, 2012

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$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25

Vertical distance is

$1-.25=.75$

- Mar 1, 2012

- 658

You calculated the vertical difference between the respective y-values of each vertex ... only problem is, the two vertices are not vertically aligned. The vertex of $y=x^2+1$ is at $(0,1)$ and the vertex of $y=x-x^2$ is $\left(\frac{1}{2},\frac{1}{4}\right)$

$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25

Vertical distance is

$1-.25=.75$

To get the idea, try the same question with these two functions ...

What is the minimum vertical distance between $y = x^2+1$ and $y = x-1$

... obviously, $y=x-1$ doesn't have a vertex.

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- #5

- Jan 31, 2012

- 2,685

Yes the word distance was ackward as opposed to difference

This was an even problem # so no answer was given so you are probably correct

- Jan 29, 2012

- 1,151

You could also use "f'(x)= 0" with f(x)= 2x^2-x+ 1. f'(x)= 4x- 1= 0. which gives x= 1/4 again.