Mathematics

Mathematics Solution

Graphing Quadratic Equations.

Quadratic Equations are graphed as parabolas. So in order to construct a parabola, we will first simplify the equation if necessary and then find the vertex before proceeding for calculating the values of 'y' to the corresponding values of 'x'.

Question 1

Y = -(x2+4)+6

Y = -(x2+8x+16) + 6 ------(Eq1)

For 'h' and 'k'

y = -x2 -8x-16 + 6

y = -x2-8x-10

y = ax2+bx+c

h = -b/2a

h = -(-8)/2(-1) = -4

k = (4ac-b2)/4a

k = (4*-1*-10 - (-8^2))/4*-1 = 6

(h,k) = (-4,6)

Since coeffient of x = -1, the parabola will be opening down, there for we will check the graphs where we can find the vertex coordinates lying on (-4,6) and parabola opening down.

Answer = C

Question 2

y = -x2 + 2x -4

here a = -1

b = 2

c = -4

Since a = -1, the parabola will be opening down.

Now to check the coordinates of vertex, we will calculated (h,k)

y = ax2+bx+c

h = -b/2a

h = -(2)/2(-1) = 1

k = (4ac-b2)/4a

k = (4*-1*-4 - (2^2))/4*-1 = -3

(h,k) = (1,-3)

The graph opening down and the vertex of parabola lying at (1,-3) is B

Answer B

Question 3

Solution

Given that h = -9.8t2+88.2t

We have to find the range of value of 't' when 'h' > 196

Therefore we have to find

-9.8t2 + 88.2t > 196

Dividing both sides by 9.8, we get

-t2 + 9t>20

-t2 + 9t -20>0

Creating factors, we get

-t2 +5t+4t-20>0

-t(t-5)+4(t-5)>0

-(t-4) (t-5) >0

(t-4) (t-5)<0

Therefore 4
Answer A

Question 4

Q. Use the remainder theorem and synthetic division to find f(k). k = -2; f(x) = 4x3- 4x2 - 5x + 23

Synthetic Division

-2

4

-4

-5

23

-8

24

-38

2

-12

19

-15

By Synthetic Division the answer is (D)

Remainder Theorem

4x2+4x+3

x-2

4x3 -4x2-5x+23

-4x3+8x2

0 4x2 -5x-23

4x2+ 8x

0 +3x +23

3x + 6

0 + 29

Q(x) = 4x2+4x+3 with remainder 29

Now we can say that 4x3 -4x2-5x+23 = (x-2) (4x2+4x+3) + 29

Q(x) = (x-2) (4x2+4x+3) + 29

Since k= -2, we put x = -2

Q(-2) = (-2-2) (4(-2)^2+4(-2)+3) + 29

Q(-2) = (-4) (16-8+3) +29

= (-32-12+29)

Q(-2)= 15

Hence the synthetic division and remainder theorem both gave the remainder to be 15

Answer is D

Question 5

Factor f(x) into linear factors given that k is a zero of f(x).

5) f(x) = x3 - 3x2 - 25x + 75 ; k = 5

Since k = 5 , therefore x= 5

X-5=0, which implies that (x-5) is one of the factors.

To find out the remaining factors, we will divide the polynomial with x-5.

X2+2x-15

x-5

x3 - 3x2 - 25x + 75

-x3 +5x2

0 2x2- 25x + 75

2x2 + 10x

0 -15x + 75

15x - 75

0 0 So x2 + 2x -15 = 0

X2 +5x-3x-15 = 0

X(x+5) - 3(x+5) = 0

(x+5) (x-3) = 0

Hence our factors are (x-5) (x+5) (x-3)

Answer C

Question 6

f(x) = x3 - 75x - 250; k = -5 (multiplicity 2)

The two factors are (x+5) and (x+5), in order to find the third factor, we will have to divide the polynomial twice by x+5

X2-5x-50

X+5

X3+0x2-75x-250

-x3-5x2

0 - 5x2-75x -250

5x2 +25x

0 -50x -250

50x +250

0 0

Again dividing x2-5x-50 by x+5

x-10

X+5

X2-5x-50

-X2-5X

0 -10X- 50

10x+ 50

0 0 Hence our factors are (x+5)^2 (x-10)

Answer is B

Question 7

Use synthetic division to ...