Download QUADRATIC FUNCTION Math Worksheet 7 Function equation is;

QUADRATIC FUNCTION
Math Worksheet 7
Function equation is;
y  ax 2  bx  c , the graph of this function is parable
a, b, c – coefficients (numbers)
To calculate the intersection with x-axis we calculate the quadratic equation using the following
Roots: x1, 2 
D  b 2  4ac
formulae: Discriminant
b D
2a
a>0
MAX/MIN
 b
b2 

,
c



4a 
 2a
a<0
y
y
c
c
1
0 x2
x1
1
x
Parable is convex, because a>0.
-
Domain = R
-
Range = c 
-
Decreasing   , 


Increasing 
b2
, 
4a
b
2a
Bounded below, not bounded above
-
b
In point x  
there is local
2a
minimum
1
x1
x
Parable is concave, because a>0.
-
Domain = R
-
Range =   , c 
-
Increasing   , 
b
, 
2a
-
1
x2 0





Decreasing 
b2
4a
b
2a
b
, 
2a
-
Bounded above, not bounded below
-
In point x  
b
there is local
2a
maximum
EXAMPLE:
Does the curve y = - x² - 4x, have maximum or minimum? Find the roots and sketch the graph.
ROOTS: For finding the roots use the formulae.
From the equation the coefficients are: a = -1; b = -4; c = 0
4  16  4 1.0 4  4
x1 =

 4
 2
2 1
Worked out by Jakubíková K.
 b  b 2  4ac 4  4
x2 =

0
 2
2a
1
Because we know that a < 0 the curve is concave, so it has the maximum point with the
 b
b2 
[x, y] To find this coordinates, we use the formulae:  , c   .
4a 
 2a
coordinates
The maximum point has coordinates [x, y]. The value is 4 for the x = -2
y
MAX [-2, 4]
4
x
0
-2
-4
EXERCISES:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
m)
n)
o)
p)
q)
y=2x² - 5x + 2
y = x + 6x + 9
y = 3x² + 7x + 5
y = 5x2-25x
y = 2x² + 3x – 2
y = - x² + 2x + 8
y = 6 - x² + 8x
y = x2 - 16
y = 2x² + 3x + 1
y = 3x² + 2x + 2
25
y = -x2 + 4
y = 5 + 2x - x²
y = - 2x² - 8x
y = x² + 2
y = x² - 4x + 4
y = x² - 2x
y = x² - x -12
Challenging exercises *
+
𝑥−2
3. 𝑥−𝑥 2 =
1−𝑥
1.
2
2−𝑥
1
5.
𝑥−2
𝑥
1
2
𝑥
=
𝑥2
2.
2 𝑥+2
𝑥
=
𝑥
6
+
3−𝑥
𝑥−2
4−𝑥
𝑥−1
𝑥−2
𝑥−3
2𝑥+1
1−𝑥
𝑥−4
6. 𝑥+1 − 𝑥+2 = 𝑥+3 − 𝑥+4
2
7. 2𝑥−𝑥 2 + 𝑥 2 +2𝑥 = 4−𝑥 2
SOLUTONS: 1. 0; 2. 0;5; 3. 0; 4. 4; 5. 0 6. 0, 
Worked out by Jakubíková K.
13
4. 3−2𝑥 + 2𝑥 2 −5𝑥+3 =
− 𝑥 2 −2𝑥 − 2−𝑥 = 0
𝑥−4
𝑥−3
3𝑥+3
− 𝑥−1
4
𝑥−2
2
5
; 7. 3
2
QUADRATIC INEQUALITIES
Let’s explain it using example:
x² - 9 ≥ 0 – we can solve this quadratic inequality either graphically or
algebraically
As the first thing calculate the roots of the inequality like with equation using formulae.
Graphically
Arrows indicate that for all values above
3 , 3 included is parabola positive and
for all values below -3 ( included ) is
parabola as well positive . Therefore solution
-3
0
3
s written below.
( - ∞ , -3 > U < 3 , ∞ )
Algebraically
x² - 9 ≥ 0
(x–3)(x+3)≥0
-3
III. Exercises
( - ).( - ) ≥ 0
3
( - ).( + ) < 0
( + ).( + ) ≥ 0
By factorising expression x² - 9 we get 2
1) x² - 4 <
zero values. Numbers 3, -3 divide
2) x² - 4x + 3 >
the number line into 3 intervals. On each
3) x² - 5x + 6 ≤
of them we do the sign test by substituting
4) 2x² + 5x - 3 >
values falling into the intervals. It is
5) x² - 2x + 1 ≤
obvious from the calculation above how it
6) 3x² + 5x -2 ≤
works . Solution is ( - ∞ , -3 > U < 3 , ∞ )
7) x² + 1 ≥ 0
8) x² + 4 < 0
9) -x² -6x – 8 ≥ 0
10) -3x² + 4x + 4 > 0
Worked out by Jakubíková K.
11) -2x² + 3x ≥ 0
3