Introduction to Quadratic equations

2. Lesson outcomes:
• Revise parabolas of basic form y= ax + b
• Sketch parabolas
• Introduce parabolic equations in standard and
turning point form
• Identify and find shape, turning point, axes of
symmetry, intercepts and asymptotes
2

3. Revision
• Functions of the general form y= ax + b are called parabolic functions,
where a and b are constants.
• The effects of a and b on y= ax + b:
b affects the vertical shift:
• b >0 , y is shifted vertically upwards by b units
• The turning point of y is above the x-axis
• b is the y –intercept
2
2
y= x
2
y= x + 1
2
Graph
has
shifted 1
unit
upwards
b=0
b=1

4. • b < 0, y is shifted vertically downwards by b units.
• The turning point of y is below the x-axis
-1
0
y = x - 1
2
y= x
2
b=0
b= - 1

6. Effects of a on the graph shape
• For a > 0; the graph of y is a “smile” and has a minimum turning
point (0;b). As the value of a becomes larger, the graph becomes
narrower.
• For a < 0; the graph of y is a “frown” and has a maximum turning
point (0;q). As the value of a becomes smaller, the graph becomes
narrower.
b >0
b =
0
b <0

7. y= x
y= x
y= 2x
y= 3x
2

8. Functions of the form
y=a(x+p)+q
p shifts the graph horizontally
• For p > 0, the graph is shifted to the left by p units.
• For p < 0, the graph is shifted to the right by p units.
The value of p also affects whether the turning point is to the left of the
y-axis (p>0) or to the right of the y-axis (p<0).
• The axis of symmetry is the line x=−p

9. q has the same effect as ‘b’
• For q >0, y shifts q units upwards
• For q < 0 y shifts q units downwards

10. Characteristics for
• The domain is {x:x∈R} because f(x)=y is defined for all x values
• The range depends on whether the value for a is positive or
negative. If a>0 we have:
• ( since a pefect square is always + )
• ( a is + )
• Thus f(x) q
• The range is therefore {y :y≥ q, y ∈ R} if a > 0. Similarly, if a < 0,
the range is {y : y ≤ q ,y ∈ R}

11. Example: DOMAIN AND RANGE
State the domain and range for g(x)= −2(x−1) + 3.
Determine the domain
The domain is {x:x∈R} because there is no value of x for which g(x) is
undefined.
Determine the range
The range of g(x) can be calculated from:
(x-1) 0
-2(x-1) 0
-2 (x-1) +3 3
g(x) 3
Therefore the range is {g(x):g(x)≤3} or in interval notation (−∞;3].

12. Intercepts
The y-intercept: let x=0.
example,
the y-intercept of g(x)= (x−1)+ 5
g(0)=(x−1) +5=(0−1) +5=6
This gives the point (0;6).
The x-intercept: let y=0.
g(x)=(x−1) +5
0= (x-1) + 5
-5 = (x-1)
which has no real solutions. Therefore, the
graph of g(x) lies above the x-axis and
does not have any x-intercepts.

13. Turning points
The turning point of the function f(x)=a(x+p) +q is determined by examining
the range of the function:
If a>0, f(x) has a minimum turning point and the range is [q;∞):
• The minimum value of f(x) is q.
• If f(x)=q, then a(x+p) =0, and therefore x=−p.
• This gives the turning point (−p;q).
If a<0, f(x) has a maximum turning point and the range is (−∞;q]:
• The maximum value of f(x) is q.
• If f(x)=q, then a(x+p) =0, and therefore x=−p.
• This gives the turning point (−p;q).
• Therefore the turning point of the quadratic function f(x)=a(x+p) +q is
(−p;q)

14. Determine the turning point of
y = 3x -6x -1
Step 1:
Y= 3 ( x - 2x ) – 1 use (–b/2) = (-2/2) = 1
Y= 3 (x - 2x +1 -1) -1 add and subtract the value next to bx and
Y= 3 (( x - 1 ) -1) -1 factorize the underlined equation
Y= 3 (x -1) -3-1 simplify
y= 3 ( x-1 ) -4
Step 2: determine the turning point (-p; q)
P= -1
q= -4
Thus the turning point is (-(-1); -4) = (1; -4)
Write the equation in the form y=a(x+p)
+q

15. Axis of symmetry
The axis of symmetry for f(x)=a(x+p) +q is the vertical line x=−p. The
axis of symmetry passes through the turning point (−p;q) and is parallel
to the y-axis.

16. Sketching graphs of the form
f(x)=a(x+p)+q
In order to sketch graphs of the form f(x)= a(x+p)
+q, we need to determine five characteristics:
• sign of a (+ or - )
• turning point (-p; q)
• y-intercept (x=0)
• x-intercept(s) (if they exist) (y=0)
• domain and range

17. Sketch the graph of y=−(1/2)(x+1)−3.
Mark the intercepts, turning point and the axis of symmetry. State the
domain and range of the function.
Examine the equation of the form y=a(x+p)+q
• We notice that a<0, therefore the graph is a “frown” and has a
maximum turning point.
• Determine the turning point (−p;q)
From the equation we know that the turning point is (-p; q) = (−1;−3).
• Determine the axis of symmetry x=−p
From the equation we know that the axis of symmetry is x=−1
• Determine the y-intercept
The y-intercept is obtained by letting x=0:
y=−(1/2)((0)+1) −3=(−1/2)−3= -
3.5
This gives the point (0;−3.5).

18. • Determine the x-intercepts
The x-intercepts are obtained by letting y=0:
0= -(1/2) (x + 1) -3
3x (-2)= (x+1)
which has no real solutions. Therefore, there are no x-intercepts
and the graph lies below the x-axis.
• Plot the points and sketch the graph.
State the domain and
range
Domain: {x:x∈R}
Range: {y:y≤−3,y∈R}

19. Sketch the graph of y=(1/2)x −4x + (7/2).
• Examine the equation of the form y=ax +bx+c
We notice that a>0, therefore the graph is a “smile” and has a
minimum turning point
• determine the turning point and the axis of symmetry
Check that the equation is in standard form and identify the
coefficients.
a=(1/2) ;b=−4; c=(7/2)
Calculate the x-value of the turning point using
x=−b/2a =−(−4 /2(1/2)) =4
Therefore the axis of symmetry is x=4.
Substitute x=4 into the original equation to obtain the
corresponding y-value.
y=−4.5
This gives the point (4;−4.5).

20. • Determine the y-intercept
The y-intercept is obtained by letting x=0:
y=(1/2)(0)−4(0)+(7/2) =7/2
This gives the point (0;7/2)
• Determine the x-intercepts
The x-intercepts are obtained by letting y=0:
0=(1/2)x −4x+(7/2) = x −8x+7=(x−1)(x−7)
Therefore x=1 or x=7. This gives the points (1;0) and (7;0)
• Plot the points and sketch the graph
Domain: {x:x∈R}
Range: {y:y ≥ −4.5 ,y∈R}