sign in sign up
<<
Home , Discriminant, Quadratic equation, Quadratic formula

Quadratic

If a>0thenthegraphofax2 is obtained by starting with the graph of x2, and then stretching or shrinking vertically by a. If a<0thenthegraphofax2 is obtained by starting with the graph of x2, then flipping it over the x-axis, and then stretching or shrinking vertically by the positive a. When a>0wesaythatthegraphof ax2 “opens up”. When a<0wesay that the graph of ax2 “opens down”. I Cit

i-a

x-ax~S

~12 *************‘s-aXiS —10.?

148 2 If a, c, d and a = 0, then the graph of a(x + c) 2 + d is obtained by If a, c, d R and a = 0, then the graph of a(x + c)2 + d is obtained by 2 R 6 2 shiftingIf a, c, the d ⇥ graphR and ofaax=⇤ 2 0,horizontally then the graph by c, and of a vertically(x + c) + byd dis. obtained (Remember by shiftingshifting the the⇥ graph graph of of axax⇤ 2 horizontallyhorizontally by by cc,, and and vertically vertically by by dd.. (Remember (Remember thatthatd>d>0meansmovingup,0meansmovingup,dc>0meansmoving0meansmoving thatleft,andd>c<0meansmovingup,0meansmovingd0meansmoving leftleft,and,andc

2 Example. Below is the parabola that is the graph of 10(x +2)2 3. Its Example. Below is the parabola that is the graph of 10(x +2)2 3. Its vertexExample. is ( 2,Below3). is the parabola that is the graph of 10(x +2) 3. Its vertexvertex is is ( (22,,3).3).

************* ************************** 149 2 2 A quadratic is a degree 2 polynomial. In other words, a qua- dratic polynomial is any polynomial of the form p(x)=ax2 + bx + c where a, b, c R and a =0. 2 6 Completing the You should memorize this : (it’s called .)

b 2 b2 ax2 + bx + c = a x + + c 2a 4a ⇣ ⌘

Let’s check that the equation is true:

b 2 b2 b b 2 b2 a x + + c = a x2 +2x + + c 2a 4a 2a 2a 4a ⇣ ⌘ ⇣ b h bi ⌘2 b2 = ax2 + a2x + a + c 2a 2a 4a b2 h ib2 = ax2 + bx + a + c 4a2 4a bh2 ib2 = ax2 + bx + + c 4a 4a = ax2 + bx + c

Graphing quadratics Complete the square to graph quadratic polynomials: If p(x)=ax2 +bx+c, b 2 b2 2 then p(x)=a x + 2a + c 4a. Therefore, the graph of p(x)=ax + bx + c 2 b is a parabola obtained by shifting the graph of ax horizontally by 2a,and 2 vertically by c b . The parabola opens up if a>0andopensdownifa<0. 4a Example. To graph 3x2 +5x 2, complete the square to find that 3x2 +5x 2equals 3(x 5)2 + 1 . To graph this polynomial, we start 6 12 with the parabola for 3x2, which opens down since 3isnegative. 150 . .

Shift the parabola for 3x2 right2 by 5 5and then up by 1 1. The result is the Shift the parabola for 3x right by6 6 and then up by1212. The result is the 22 2 5 1 22 graphgraphShift for for the3 parabola3xx +5+5xx for2.2. 3 Noticex Noticeright that that by the6 theand graph graph then looks looks up by like like12 the. the The graph graph result of of is33 thexx,, 2 55 11 2 exceptgraphexcept for that that3 its itsx vertex+5 vertexx is is2. the the Notice point point that ( (6,, the12).). graph looks like the graph of 3x , 56 121 except that its vertex is the point (6, 12).

************************** *************

4 151 4 The discriminant of ax2 + bx + c is defined to be the number b2 4ac. How many roots? If p(x)=ax2 + bx + c, then the following chart shows how the discriminant of p(x)determineshowmanyrootsp(x)has:

b2 4ac number of roots > 0 2

=0 1

< 0 0

Example. Suppose p(x)= 2x2 +3x 1. Because 32 4( 2)( 1) = 9 8=1ispositive,p(x)= 2x2 +3x 1hastworoots.

Why the discriminant of a quadratic polynomial tells us about the number of roots of the polynomial, and why the information from the above chart is true, will be explained in lectures.

*************

152 Finding roots If ax2 + bx + c has at least one root – which is the same as saying that b2 4ac 0–thenthereisaformulathattellsuswhatthoserootsare.

If p(x)=ax2 + bx + c with a =0andifb2 4ac 0, then the roots of6 p(x)are

b + pb2 4ac b pb2 4ac and 2a 2a

Notice in the quadratic formula, that we need a =0tomakesurethatwe are not dividing by 0, and we need b2 4ac 0tomakesurethatwearen’t6 taking the of a . Also recall that if ax2 + bx + c has only one root, then b2 4ac =0.That means the two roots from the quadratic formula are really the same root. It’s a good exercise in to check that the is true. To check that it’s true, you need to check that b + pb2 4ac p =0 2a and that ⇣ ⌘ b pb2 4ac p =0 2a ⇣ ⌘ Example. We checked above that p(x)= 2x2+3x 1has2roots,because its discriminant equalled 1. The quadratic formula tells us that those roots equal 3+p1 3+1 2 1 = = = 2( 2) 4 4 2 and 3 p1 4 = =1 2( 2) 4

153 Exercises For each of the quadratic polynomials in problems #1-6:

Complete the square. • 2 2 That means rewrite ax2 + bx + c as a x + b + c b . 2a 4a What’s the vertex of the corresponding⇣ parabola?⌘ • The vertex of the parabola for a(x + c)2 + d is the point ( c, d). Is its parabola opening up, or opening down? • The parabola opens up if the leading coecient of the quadratic polynomial is positive. The parabola opens down if the leading coecient is negative. What’s its discriminant? • The discriminant of ax2 + bx + c is b2 4ac. How many roots does it have? • There are two roots if the discriminant is positive, one root if the discriminant equals 0, and zero roots if the discriminant is negative. What are its roots (if it has any)? • 2 b+pb2 4ac b pb2 4ac If ax + bx + c has roots, they are 2a and 2a . Match its graph with one of the lettered graphs on the next page. •

1.) 2x2 2x +12 2.) x2 +2x +1

3.) 3x2 9x +6 4.) 4x2 +16x 19 5.) x2 +2x 1 6.) 3x2 +6x +5

154 I) z~j) I) a 1) z~j) I) a 1) -‘4

I) —I I)I) —I I) 3) z~j) 2 I) a 1) I) 3) z~j) 2 I) a 1) (z,.-3) -‘4 —3” :3’,s’) I) —I (z,.-3) —3” —I I) I) s):3’, 1+ I) C) z~j) I) 3) —I z~j) I) A.)I) B.)1) 2 s) a 1+ —z I) z~j) s’)1) —I C) (-1,2) a - z) I) 3) z~j)I) a 1) 2 z~j) I) a 1) I) a z~j) 1) (-1,2) s’) 1) -‘4 (z,.-3) I) a -‘4 —I —3” :3’,s’) —I —z -‘4 I) s’) —I - z) (z,.-3)—I I) —3” s’) I) —I I) —I I) s):3’, 1+ —I 3) I) —z C) —I I) I) s’) —I 2 3) C.)-—Iz) D.) I) I) z~j)2 —z 3) s) s’) I) 1+ —z—I 1) I) —I - z) I) 2 C) (-1,2) 3) a - z) z~j) 2 I) I) —z 3) a z~j)2 1) 3) I) - z) 1) I) a z~j)2 (-1,2)(z,.-3) s’) (z,.-3) —z—3” —I - 1) —3”I) a -‘4 z) s’) :3’,s’) —z (z,.-3) —I —3”—I (z,.-3) :3’,s’) - z) -‘4 —I (z,.-3) —3” :3’,s’) I) —I s’) —3”—I —I s):3’, 1+ (z,.-3) :3’, I)I) —3” I)C) s) 1+ —zs’) —I :3’, C) —I s) 1+ E.)3) I)- z) F.) —I z~j) s)I) 1+ C) —z —z 2 s’) —I I)C) (-1,2) 1) - s) 1+ —z - z) z) a —I s’) —I I)C)3) (-1,2) 2 - z) s) 1+ —z —z I) z~j) s’) —I 3) C) (-1,2) 2 - z) - z) 1) I) I) a (-1,2) s’) 3) (-1,2) —z 2 —I (z,.-3) s’) - z) -‘4 —I s’) —z (-1,2) —3” s’) —I s’) —I —z - z) (z,.-3) —I :3’,s’) —3” - —I I) s’) z) —zs’) (z,.-3) —I —I —I —3” - z) :3’,s’) s’) —I —I 155 I) (z,.-3) :3’, 1+—zs’) s’) —I s) —I —3” —zs’) - z) 3) C) —I —I :3’, —z I) 2 - z) —z s) 1+ —z s’) —I - z) C) - z) s) s’) 1+ —z - z) —I C) (-1,2)I) —z —z - z) 3) 2 —z - z) s) - z) 1+ —z s’) —I C) (-1,2) - - z) —z z) —z (-1,2) (z,.-3) - - z) z) s’) —z —3” - z) —I—z (-1,2) :3’, - z) —z s’) s’)s’) —I—I —I (z,.-3) - z) s’) —z —3” s’) —I —I - z) :3’, s) 1+ C) s’) —I —z s’) —I - z) —z—z s) 1+ —z s’) s’) —I C) (-1,2) —I - - -z)z) z) —z —z - z) - z) (-1,2) —z s’) —z - z) —I - z) s’) —z —I —z - - z) z)

s’) —I —z s’) —I - z) —z

- z) —z

- z) —z

- z) 7.) Suppose you shoot a feather straight up into the air, and that t is the time measured in seconds that follow after you shoot the feather into the air. If the height of the feather at time t is given by 2t2 +20t feet, then what is the maximum height that the feather reaches? How many seconds does it take for the feather to reach its maximum height?

8.) Let’s say you make cogs for a living. After accounting for the cost of building materials, you earn a profit of x2 10x +45centsonthex-th cog that you make. Which cog do you earn the least amount of profit for making? How much profit do you earn for that cog?

Solve for x in the given in #9-11.

x2 + x +1 x2 +3 3x 5 9.) =1 10.) =1 11.) = 2 2x +1 x2 +2x +5 x2

156