Minimum and Maximum points on a curve, etc

N

nil101

New member
Joined
Oct 16, 2005
Messages
37
I need some help with this problem please.
Use the method of completing the square to find the axis of symmetry and the maximum or minimum point of the curve
y=2+x−x2\displaystyle y = 2 + x - x^2
and sketch the curve showing it min or max, and its intercepts with the x and y axes.


This is how far i've got
\(\displaystyle {
y = - (x^2 - x - 2) \cr
= - (x^2 - x + \frac{{x^2 }}{4} - 2 - \frac{{x^2 }}{4}) \cr
= - \left[ {(x - {\textstyle{x \over 2}})^2 - \frac{{8 - x^2 }}{4}} \right] \cr}\)

I make the min or max point
(x2,8−x24)\displaystyle \left( {\frac{x}{2},\frac{{8 - x^2 }}{4}} \right)
And the intercept on the y-axis -2

I’m not sure if the above co-ordinate is the min or the max point and I don’t think the y coordinate is right. So I’m having trouble working out the x-axis intercepts or sketching the curve.

Can you help please?
 
When you complete the square you arrive at the standard equation of a parabola.

y=a(x−h)2+k\displaystyle y=a(x-h)^{2}+k

Where (h,k) is the coordinates of the vertex.

−x2+x+2\displaystyle {-}x^{2}+x+2

−(x2−x)=−2\displaystyle {-}(x^{2}-x)=-2

−(x2−x+14)=−2−14\displaystyle {-}(x^{2}-x+\frac{1}{4})={-}2-\frac{1}{4}

−(x−12)2=−94\displaystyle {-}(x-\frac{1}{2})^{2}={-}\frac{9}{4}

y=−(x−12)2+94\displaystyle y={-}(x-\frac{1}{2})^{2}+\frac{9}{4}

To find the x-intercept set y=0 and solve for x.

To find the y-intercept set x=0 and solve for y.

Also, you can find the vertex by using −b2a\displaystyle \frac{-b}{2a}

from your original equation.
 
nil101 said:
I need some help with this problem please.
Use the method of completing the square to find the axis of symmetry and the maximum or minimum point of the curve
y=2+x−x2\displaystyle y = 2 + x - x^2
and sketch the curve showing it min or max, and its intercepts with the x and y axes.


This is how far i've got
\(\displaystyle {
y = - (x^2 - x - 2) \cr
= - (x^2 - x + \frac{{x^2 }}{4} - 2 - \frac{{x^2 }}{4}) \cr
= - \left[ {(x - {\textstyle{x \over 2}})^2 - \frac{{8 - x^2 }}{4}} \right] \cr}\)

I make the min or max point
(x2,8−x24)\displaystyle \left( {\frac{x}{2},\frac{{8 - x^2 }}{4}} \right)
And the intercept on the y-axis -2

I’m not sure if the above co-ordinate is the min or the max point and I don’t think the y coordinate is right. So I’m having trouble working out the x-axis intercepts or sketching the curve.

Can you help please?
Click to expand...

Hi!


First of all, to find the y-intercept, substitute 0 for x in the original equation:
y = -x<SUB>2</SUB> + x + 2
y = - (0)<SUB>2</SUB> + 0 + 2

Solve this for y...that will be the y-intercept.

To find the x-intercept(s), subsititute 0 for y....
0 = -x<SUB>2</SUB> + x + 2
Multiply both sides by -1:
0 = x<SUB>2</SUB> - x - 2
Factor the right side:
0 = (x - 2)(x + 1)
Set each factor equal to 0 and solve for x....this will tell you where the graph crosses the x axis.

If an equation is of the form
y = ax<SUP>2</SUP> + bx + c
then its graph is a parabola. If the coefficent of x<SUP>2</SUP> is greater than or equal to 1, the parabola opens upwards, and the lowest point (minimum value of the function) occurs at the vertex. If the coefficient of x<SUP>2</SUP> is negative, then the parabola opens downward....looks like an upside-down U....and the highest point (maximum value of the function), also occurs at the vertex.

It looks like you may have been trying to complete the square, which is a good way to put the function in vertex form,
y = a(x - h)<SUP>2</SUP> + k

Then, the vertex of the parabola is the maximum or the minimum value of the function, depending on whether the parabola opens up or down.

Your function is

y = 2 + x - x<SUP>2</SUP>

Let's rearrange this:
y = - x<SUP>2</SUP> + x + 2

Completing the square is much easier if the coefficent of the x<SUP>2</SUP> term is 1. So, let's multiply both sides of the equation by -1:

-y = x<SUP>2</SUP> - x - 2

Next, let's get the terms containing x by themselves. Add 2 to both sides of the equation:

-y + 2 = x<SUP>2</SUP> - x

Divide the coefficient of x by 2, square it, and add the result to both sides of the equation. -1 divided by 2 is -1/2, and (-1/2)<SUP>2</SUP> = 1/4. Add 1/4 to both sides of the equation:

-y + 2 + (1/4) = x<SUP>2</SUP> - x + (1/4)

-y + (9/4) = (x - 1/2)<SUP>2</SUP>

-y = (x - 1/2)<SUP>2</SUP> - (9/4)

Now, multiply both sides by -1 to get 1y, or just y, on the left side:

y = -1(x - 1/2)<SUP>2</SUP> + 9/4

This equation is now in "vertex form." The vertex of the parabola is at (1/2, 9/4).

To find the maximum value of the function, substitute 9/4 for x in the original equation.

I hope this helps you.

(edited to correct careless sign errors)
 
Hello, nil101!

Wow . . . everyone beat me to the punch!
    \displaystyle \;\;Well, I'm not going to delete all this . . .

Use the method of completing the square to find the axis of symmetry and the maximum or minimum point of the curve
y=2+x−x2\displaystyle y = 2 + x - x^2
and sketch the curve showing it min or max, and its intercepts with the x and y axes.

This is how far i've got
y = −(x2 − x − 2)\displaystyle y\:=\: -(x^2\,-\,x\,-\,2)
    = −(x2 − x + x24−2−x24) \displaystyle \;\; =\:-(x^2\,-\,x\,+\,\frac{{x^2 }}{4} - 2 - \frac{{x^2 }}{4})\: <--- This is wrong (a simple error)
Click to expand...
We take half of the coefficient of x . . . and square it.
    \displaystyle \;\;So:  12 × 1 = 12    ⇒    (12)2 = 14\displaystyle \,\frac{1}{2}\,\times\,1\:=\:\frac{1}{2}\;\;\Rightarrow\;\;\left(\frac{1}{2}\right)^2\:=\:\frac{1}{4}

Then we have:  y  =  −(x2 − x + 14 − 2 − 14)  =  −(x2 − x + 14 − 94)  =  −(x2 − x + 14) + 94\displaystyle \:y\;=\;-\left(x^2\,-\,x\,+\,\frac{1}{4}\,-\,2\,-\,\frac{1}{4}\right)\;=\;-\left(x^2\,-\,x\,+\,\frac{1}{4}\,-\,\frac{9}{4}\right)\;=\;-\left(x^2\,-\,x\,+\,\frac{1}{4}\right)\,+\,\frac{9}{4}

Hence:  y  =  −(x − 12)2 + 94\displaystyle \:y\;=\;-\left(x\,-\,\frac{1}{2}\right)^2\,+\,\frac{9}{4}

The axis of symmetry is the vertical line:  x = 12\displaystyle \,x\,=\,\frac{1}{2}

The vertex is:  (12, 94)\displaystyle \,\left(\frac{1}{2},\,\frac{9}{4}\right)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For x-intercepts, let y = 0\displaystyle y\,=\,0
    0  =  −(x − 12)2 + 94    ⇒    (x − 12)2  =  94\displaystyle \;\;0\;=\;-\left(x\,-\,\frac{1}{2}\right)^2\,+\,\frac{9}{4}\;\;\Rightarrow\;\;\left(x\,-\,\frac{1}{2}\right)^2\;=\;\frac{9}{4}

Take square roots:  x − 12  =  ±94  =  ±32\displaystyle \,x\,-\,\frac{1}{2}\;=\;\pm\sqrt{\frac{9}{4}}\;=\;\pm\frac{3}{2}

Then:  x  =  12 ±32  =  2, −1\displaystyle \,x\;=\;\frac{1}{2}\,\pm\frac{3}{2}\;=\;2,\,-1

The x-intercepts are:  (2, 0)\displaystyle \,(2,\,0) and (−1, 0)\displaystyle (-1,\,0).


For y-intercepts, let x = 0\displaystyle x\,=\,0
    y  =  −(0 − 12)2 + 94  =  −14 + 94  =  2\displaystyle \;\;y\;=\;-\left(0\,-\,\frac{1}{2}\right)^2\,+\,\frac{9}{4}\;=\;-\frac{1}{4}\,+\,\frac{9}{4}\;=\;2

The y-intercept is:  (0,2)\displaystyle \,(0,2)


The graph looks like this:
Code: 
                |   (1/2,9/4)
                |      *o*
                |  *         *
               2o               * 
           -1 * |                 * 2
        -----o--+------------------o---
                |
            *   |                   *
 
Wow guys, thanks so much for your help with this question. I really am grateful for the good explanation.

I need a more help to answer this part of the qusetion:

Find the equation of the tangent to the curve y=2 + x - x^2 at the point where it crosses the y axis.

 
When it crosses the y-axis, x will be 0. When you enter x=0 into your equation,

what does y equal?. Find the slope by differentiating and then finding the slope at

x=0. That will be m.


Sub these values into y=mx+b, solve for b, and you have your equation.

Do you have a graphing calculator?. Graph them both and see if the line is tangent

to the parabola where it crosses the y-axis.
 
You must log in or register to reply here.
Top