# [SOLVED]s8.3.7.6 minimum vertical distance

#### karush

##### Well-known member
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

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

#### skeeter

##### Well-known member
MHB Math Helper
S8.3.7.6. What is the minimum vertical distance between the parabolas
$$y = x^2+1 \textit{and } y = x- x^2$$

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

Last edited:

#### karush

##### Well-known member
$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$

#### skeeter

##### Well-known member
MHB Math Helper
$(x^2+1)'=2x$. Then 2x=0 so x=0
$(x-x^2)'=1-2x$ then 1-2x=0 so x=.5

(0)^2+1=1 And (.5)-(.5)^2=.25
Vertical distance is
$1-.25=.75$
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 $\left(\frac{1}{2},\frac{1}{4}\right)$

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.

#### karush

##### Well-known member
Ok well this was from exercises in doing min/max problems where all the examples were solved by $f'(x)=0$

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

#### HallsofIvy

##### Well-known member
MHB Math Helper
You are still ignoring skeeter's first response! No, this not asking for the distance between vertices, it is asking for the minimum of $$\displaystyle |y_1- y_2|= |x^2+ 1- x+ x^2|= |2x^2- x+ 1|$$. That (ignoring the absolute value) is a parabola, $$\displaystyle 2(x^2- x/2+ 1/16- 1/16)+ 1= 2(x- 1/4)^2+ 15/16$$. That's always positive so we can drop the absolute value. It has a minimum value of 15/16 when x= 1/4.

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.