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Instructions and examples on how to find the vertex of a quadratic function, which is the point where the parabola reaches its maximum or minimum value. The vertex can be found by using the formula x = โb/2a, where a and b are coefficients of the quadratic equation ax^2 + bx + c. The document also explains how to find the y-coordinate of the vertex by substituting the x-coordinate back into the quadratic equation.
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As shown in the previous set of notes, the vertex of a quadratic function
provides information about many features of that function and its graph
(increasing/decreasing intervals, line of symmetry, maximum/minimum
function value, range). However when a quadratic function is expressed in
polynomial form
2
, the vertex of the quadratic
function is not obvious. One way to find the vertex of a quadratic function
that is in polynomial form is to use the formula ๐ฅ =
โ๐
2 ๐
to find the
๐ฅ-coordinate of the vertex. Once you have the ๐ฅ-coordinate, you can find
the ๐ฆ-coordinate by replacing ๐ฅ with
โ๐
2 ๐
in the quadratic function
โ๐
2 ๐
)). Just like the quadratic formula that weโve worked with before,
the formula ๐ฅ =
โ๐
2 ๐
is also derived by completing the square (more on this
later).
Example 1 : Find the vertex of the quadratic function
2
2
Example 2 : Find the vertex of the following quadratic functions.
a. ๐(๐ฅ) = 5 ๐ฅ
2
2
b.
c. โ
2
โ 6 ๐ฅ โ 5 d. ๐
3
4
2
All the quadratic functions weโve seen so far have been in polynomial
form
2
. However if a quadratic function is not
given in polynomial form, it can be converted to polynomial form by
using polynomial multiplication and combining like terms. This is what
weโll do on the next example.
๐ฅ =
โ๐
2 ๐
๐ (
4
3
) = 9 (
4
3
)
2
โ 24 (
4
3
) + 16
๐ฅ =
โ
( โ 24
)
2 ( 9 )
๐ (
4
3
) = 9 (
16
9
) โ 32 + 16
๐ฅ =
24
18
๐ (
4
3
) = 16 โ 32 + 16
๐ฅ =
4
3
๐ (
4
3
) = 0
๐ฅ =
4
3
๐
๐
๐ฅ =
โ๐
2 ๐
โ(โ 1 ) = โ 3 (โ 1 )
2
โ 6 (โ 1 ) โ 5
๐ฅ =
โ
( โ 6
)
2 (โ 3 )
โ
( โ 1
) = โ 3
( 1
)
๐ฅ =
6
โ 6
โ(โ 1 ) = โ 3 + 6 โ 5
๐ฅ = โ 1 โ(โ 1 ) = โ 2
๐ฅ =
4
3
Example 4 : Find the vertex of the following quadratic functions, and state
whether the vertex is the maximum or minimum point on the parabola.
a. ๐
2
โ 12 ๐ฅ + 21 b. ๐
โ๐
2 ๐
2
โ
( โ 12
)
2
( 1
)
2
12
2
โ๐
2 ๐
โ
( โ 12
)
2
( โ 3
)
2
12
โ 6
Example 5 : Find the vertex and the zeros for each of the following
quadratic functions.
a. ๐
b. ๐
2
2
2
โ
( 6
)
2
( โ 1
)
โ
( โ 6
)
2
( โ 3
)
โ 6
โ 2
6
โ 6
2
Since ๐
( ๐ฅ
) is already in polynomial form,
I do not need to change it in any way to
find the vertex, like I have to do with ๐(๐ฅ)
in part a.
Example 7: Find the vertex and the range for each of the following
quadratic functions.
a. ๐
b. ๐
2
2
โ 2
2
( 1
)
2
โ 2
2
โ(โ 8 )
2
( โ 2
)
8
โ 4
2
In order to find the range of each function, I simply need to use the ๐ฆ-
coordinate of each vertex and determine whether that is the largest
output or the smallest output. To do so, I use the leading coefficient
of each function to determine whether the graph is opening up or
down.
The leading coefficient of The leading coefficient of
๐(๐ฅ) is โ 2 , so the graph of ๐(๐ฅ) is 1 , so the graph of ๐(๐ฅ)
๐(๐ฅ) is opening down. That is opening up. That means
means 50 is the largest output โ 9 is the smallest output
Example 8: Find the vertex and the increasing/decreasing intervals for
each of the following quadratic functions.
a. ๐(๐ฅ) = โ
1
2
( 3 โ ๐ฅ)(๐ฅ + 2 ) b. โ(๐ฅ) =
2
5
2
12
5
23
5
1
2
2
โ(โ
12
5
)
2 (
2
5
)
1
2
2
1
2
12
5
4
5
โ(โ
1
2
)
2 (
1
2
)
12
4
1
2
1
1
2
2
5
2
12
5
23
5
1
2
1
2
1
2
1
2
2
5
36
5
23
5
1
2
5
2
5
2
18
5
36
5
23
5
25
8
๐
๐
๐๐
๐