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Graphs

A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are

numbers with a not equal to zero.

The graph of a quadratic function is a curve called a parabola. Parabolas may open

upward or downward and vary in "width" or "steepness", but they all have the same

basic "U" shape. The picture below shows three graphs, and they are all parabolas.

All parabolas are symmetric with respect to a line called the axis of symmetry. A

parabola intersects its ais of symmetry at a point called the vertex of the parabola.

!ou now that two points determine a line. This means that if you are given any two

points in the plane, then there is one and only one line that contains both points. A

#iven three points in the plane that have different first coordinates and do not lie on a

line, there is eactly one quadratic function f whose graph contains all three points.

The applet below illustrates this fact. The graph contains three points and a parabola

that goes through all three. The corresponding function is shown in the tet bo below

the graph. \$f you drag any of the points, then the function and parabola are updated.

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%any quadratic functions can be graphed easily by hand using the techniques of

stretching&shrining and shifting 'translation( the parabola y ) * . '+ee the section

on manipulating graphs.(

ample -.

+etch the graph of y ) *&*. +tarting with the graph of y ) *, we shrin by a factor of

one half. This means that for each point on the graph of y ) *, we draw a new point

that is one half of the way from the ais to that point.

ample *.

+etch the graph of y ) '  /(0*  1. 2e start with the graph of y ) * , shift / units

right, then 1 units down.

http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/functions/manipulate/manipulate.html http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/functions/manipulate/manipulate.html

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Exercise 13

'a( +etch the graph of y ) ' 4 *(*  5. Answer

'b( +etch the graph of y ) '  1(* 4 5. Answer

Standard orm

The functions in parts 'a( and 'b( of ercise - are eamples of quadratic functions

in standard form. 2hen a quadratic function is in standard form, then it is easy to

setch its graph by reflecting, shifting, and stretching&shrining the parabola y )  *.

The quadratic function f'( ) a'  h(* 4 , a not equal to zero, is said to be

in standard form. \$f a is positive, the graph opens upward, and if a is negative, then it opens downward. The line of symmetry is the vertical line  ) h, and the vertex is

the point 'h,(.

Any quadratic function can be rewritten in standard form by completin! the square.

'+ee the section on solving equations algebraically to review completing the square.(

The steps that we use in this section for completing the square will loo a little

different, because our chief goal here is not solving an equation.

6ote that when a quadratic function is in standard form it is also easy to find its zeros

by the square root principle.

ample 5.

2rite the function f'( ) *  7 4 8 in standard form. +etch the graph of f and find

its zeros and verte.

f'( ) *  7 4 8.

) '*  7 (4 8. #roup the * and  terms and then complete the square on these

terms.

) '*  7 4 9  9( 4 8.

2e need to add 9 because it is the square of one half the coefficient of , '7&*(* ) 9.

2hen we were solving an equation we simply added 9 to both sides of the equation.

\$n this setting we add and subtract 9 so that we do not change the function.

http://ans1a%28%29/ http://ans1b%28%29/ http://jump%28%27/#Content') http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/Izs/asolve/asolve.html#compsqr http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/Izs/asolve/asolve.html#compsqr http://ans1a%28%29/ http://ans1b%28%29/ http://jump%28%27/#Content') http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/Izs/asolve/asolve.html#compsqr

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) '*  7 4 9(  9 4 8. 2e see that *  7 4 9 is a perfect square, namely '  5(*.

f'( ) '  5(*  *. This is standard form.

:rom this result, one easily finds the vertex of the graph of f is '5, *(.

To find the zeros of f, we set f equal to ; and solve for .

'  5(*  * ) ;.

'  5(* ) *.

'  5( ) < sqrt'*(.

 ) 5 < sqrt'*(.

To setch the graph of f we shift the graph of y ) * three units to the right and two

units down.

\$f the coefficient of * is not -, then we must factor this coefficient from the * and 

terms before proceeding.

ample /.

2rite f'( ) ** 4 * 4 5 in standard form and find the verte of the graph of f.

f'( ) ** 4 * 4 5.

) '** 4 *( 4 5.

) *'*  ( 4 5.

) *'*   4 -&/  -&/( 4 5.

2e add and subtract -&/, because '-&*(* ) -&/, and - is the coefficient of .

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) *'*   4 -&/( *'-&/( 4 5.

6ote that everything in the parentheses is multiplied by *, so when we remove -&/

from the parentheses, we must multiply it by *.

) *'  -&*(* 4 -&* 4 5.

) *'  -&*(* 4 8&*.

The verte is the point '-&*, 8&*(. +ince the graph opens downward '* = ;(, the verte

is the highest point on the graph.

Exercise 23

2rite f'( ) 5* 4 -* 4 > in standard form. +etch the graph of f ,find its verte, and

find the zeros of f. Answer

"lternate method of findin! the vertex

\$n some cases completing the square is not the easiest way to find the verte of a

parabola. \$f the graph of a quadratic function has two intercepts, then the line of

symmetry is the vertical line through the midpoint of the intercepts.

The intercepts of the graph above are at 1 and 5. The line of symmetry goes

through -, which is the average of 1 and 5. '1 4 5(&* ) *&* ) -. ?nce we now that

the line of symmetry is  ) -, then we now the first coordinate of the verte is -. The second coordinate of the verte can be found by evaluating the function at  ) -.

ample 1.

:ind the verte of the graph of f'( ) ' 4 9('  1(.

http://ans2%28%29/ http://ans2%28%29/

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+ince the formula for f is factored, it is easy to find the zeros3 9 and 1.

The average of the zeros is '9 4 1(&* ) /&* ) *. +o, the line of symmetry is  ) *

and the first coordinate of the verte is *.

The second coordinate of the verte is f'*( ) '* 4 9('*  1( ) 8@'8( ) /9.

Therefore, the verte of the graph of f is '*, /9(.

"pplications

ample 7.

A rancher has 7;; meters of fence to enclose a rectangular corral with another fence dividing it in the middle as in the diagram below.

As indicated in the diagram, the four horizontal sections of fence will each be 

meters long and the three vertical sections will each be y meters long.

The ranchers goal is to use all of the fence and enclose the lar!est possible area.

The two rectangles each have area y, so we have

total area# A ) *y.

There is not much we can do with the quantity A while it is epressed as a product of

two variables. Bowever, the fact that we have only -*;; meters of fence available

leads to an equation that  and y must satisfy.

5y 4 / ) -*;;.

5y ) -*;;  /.

y ) /;;  /&5.

http://jump%28%27/#Content') http://jump%28%27/#Content')

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2e now have y epressed as a function of , and we can substitute this epression for

y in the formula for total area A.

A ) *y ) * '/;; /&5(.

2e need to find the value of  that maes A as large as possible. A is a quadratic function of , and the graph opens downward, so the highest point on the graph of A is

the verte. +ince A is factored, the easiest way to find the verte is to find the 

intercepts and average.

* '/;; /&5( ) ;.

* ) ; or /;; /&5 ) ;.

 ) ; or /;; ) /&5.

 ) ; or -*;; ) /.

 ) ; or 5;; ) .

Therefore, the line of symmetry of the graph of A is  ) -1;, the average of ; and 5;;.

6ow that we now the value of  corresponding to the largest area, we can