MPM 2D0
Unit 4: Graphing Quadratics REVIEW
Graphing Quadratics
 Base function is y = x2.
 From (0, 0) go: over 1 up 1, over 2 up 4, over 3 up 9, over 4 up 16...
 Terms: Quadratic Function, Parabola, Axis of Symmetry, Vertex, Maximum/Minimum
Translations of y = x2





y  a x  h   k
2
a = reflection if negative, vertical stretch if a > 1, vertical compression if 0 < a < 1
h = horizontal translation left if h > 0 (positive), right if h < 0 (negative)
k = vertical translation up if k > 0 (positive), down if k < 0 (negative)
Draw base graph, do stretch/compression and reflection, move each point to new vertex OR
start at new vertex and build graph
Example 1: State the transformations for each of the following, then graph using transformations.
Complete the chart provided.
b) y 
a) y  x  4
2
Down 4
1
x  22
2
c) y  2 x  6   8
2
Vertical Compression by
a factor of ½, right 2
y = x2
Reflection in the x-axis,
Vertical Stretch by a factor
of 2, left 6, up 8
y = x2
y = –2(x + 6)2 + 8
y = x2 – 4
y = ½ x2
y = x2
y = ½ (x – 2)2
y = –2x2
Property
Vertex
Axis of Symmetry
Stretch or compression
factor relative to y  x 2
Direction of opening
Values x may take
Values y may take
a
(0, –4)
x=0
b
(2, 0)
x=2
C
(–6, 8)
x = –6
None
½
2 (reflection)
Up
Set of real numbers
y ≥ –4
Up
Set of real numbers
y≥0
Down
Set of real numbers
y≤8
Graphing Quadratics With Roots
 y  a( x  r )( x  s)
 The x-intercepts (or zeros, roots) are r and s.
MPM 2D0





Find x-intercepts by setting y = 0, then solving using factoring or quadratic formula.
Find y-intercept by setting x = 0 and solving for y (or is constant term in standard form).
Find x-value of vertex by adding the two x-intercepts and dividing by two (half-way point).
Find y-value of vertex by substituting x-value of vertex into original equation.
Plot points and join with a smooth curve.
Example 2: Sketch y  2 x  4 x  6 using the x-intercepts, y-intercept, and vertex.
2
x-intercepts
Vertex
Set y = 0
x-value
0  2 x  4 x  6
 3 1
2
2

2
 1
2


0  2 x 2  2 x  3
0  2 x  3 x  1
x  3  0 x 1  0
x  3
x 1
y-value
y-intercept
y  2 x 2  4 x  6
y6
y  2(1) 2  4(1)  6
y  2  4  6
y 8
Determine the Equation From a Graph
 Forms of quadratic functions:
o Vertex Form: y  a x  h   k
o Intercept Form: y  a( x  r )( x  s)
2
o Standard Form: y  ax  bx  c
2



Identify the vertex from the graph. Substitute into y  a x  h   k .
2
Count to identify the a-value, OR pick another point and substitute into y  a x  h   k and
solve for "a"
Convert equation to standard form if required.
2
Example 3: Determine an equation for the given parabola, in
vertex form.
 3  9a
y  a x  h   k
2
y  ax  3  4
2
Use (0, 1):
1  a0  3  4
1  4  9a
2
3
a
9
1
 a
3
y 
1
x  32  4
3
MPM 2D0
Completing the Square
 Used to convert equations in standard form to vertex form
 Group x-terms.
 Factor out a-value from x-terms (not constant term)
 Divide the coefficient of the middle term by 2, square it, then add and subtract that number
inside the brackets.
 Remove the subtracted term from the brackets. On the way out of the brackets, it is multiplied
by the a-value.
 Write the perfect square trinomial as the square of a binomial.
Example 4: Convert each of the following to vertex form.
a) y  x 2  6 x  3
b) y   x 2  8 x  11

y  x
y  x

y  x2  6x  3

 6x  9  9  3
2
 6x  9  9  3

y   x  3  6
2
2



y   x 2  8 x  11
y   3 x 2  12 x  9
y   x 2 8 x  11
y  3 x 2 4 x  9

y  x
y  x

2
c) y  3 x  12 x  9


2
 8 x  16  16  11
2
 8 x  16  16  11


y  3x
y  3x


2
 4x  4  4  9
2
 4 x  4  12  9

y  3 x  2   3
y   x  4   5
2
2
Word Problems
 Look for words like "maximum/minimum." You need the vertex, so you will have to complete
the square if the equation is in standard form.
 You will also need to find points like the y-intercept and the x-intercepts.
Example 5: The height, h metres, of a batted baseball as a function of the time, t seconds, since the
ball was hit can be modeled by the function h  2.1t  10.08t  0.904
a) What was the maximum height of the ball?
b) What was its height when it was hit, to the nearest tenth of a metre?
c) How many seconds after it was hit did the ball hit the ground, to the nearest tenth of a second?
d) What was the height of the ball, to the nearest tenth of a meter, 1s after it was hit?
2
a) h  2.1t 2  10.08t  0.904

b) y-int = 0.904
The height when it was hit was 0.9 m.

  2.1t 2  10.08t  0.904

 2.1t
 2.1t

 2.1 t 2  4.8t  0.904
c) Set h = 0 and solve for t.

2
 4.8t  5.76  5.76  0.904
2
 4.8t  5.76  12.096  0.904

 2.1t  2.4   13
2
Vertex: (2.4, 13)
t
h
Therefore, the maximum height was 13 m.
0  2.1t 2  10.08t  0.904
 10.08  10.082  4(2.1)(0.904 )
t
2(2.1)
 10.08  109 .2
 4 .2
t  0.1
t  4 .9

inadmissib le
Therefore, the ball was in the air for 4.9 s.
MPM 2D0
d) Set t = 1, and solve for h
h  2.1t 2  10.08t  0.904
 2.1(1) 2  10.08(1)  0.904
 2.1  10.08  0.904
Therefore, the ball was 8.9 m high 1 second after it was hit.
 8.884
Finite Differences
 If the first differences are constant, the function is linear
 If the second differences are constant, the function is quadratic
 If neither the first nor second differences are constant, the function is neither linear nor
quadratic
Example 6: Use finite differences to determine whether the function is linear, quadratic, or neither.
x
0
1
2
3
4
y
–5
–6
–1
10
27
1st Diff
2nd Diff
–6 – (–5) = –1
–1 – (–6) = 5
10 – (–1) = 11
27 – 10 = 17
5 – (–1) = 6
11 – 5 = 6
17 – 11 = 6
Since the 2nd differences are the same, the function is quadratic.