# Quadratic Function

See how to solve a quadratic function.

17 examples and their solutions.

## Quadratic Function: Opened Upward, Downward

### Formula

y = ax

a: (+)a: (-)

a: (+) → opened upward^{2}+ bx + ca: (+)a: (-)

a: (-) → opened downward

The shape of a quadratic function,

the U shape,

is called a parabola.

### Example

y = x

Upward / downward?

Solution ^{2}+ 2x + 5Upward / downward?

y = x

→ Opened upward

^{2}+ 2x + 5→ Opened upward

Close

### Example

y = -3x

Upward / downward?

Solution ^{2}+ x - 8Upward / downward?

y = -3x

→ Opened downward

^{2}+ x - 8→ Opened downward

Close

## Quadratic Function: Axis of Symmetry

### Formula

y = ax

x = -b2a

The axis of symmetry is the line^{2}+ bx + cx = -b2a

that cuts the graph

into two symmetric pieces.

### Example

y = x

Axis of symmetry?

Solution ^{2}+ 4x - 9Axis of symmetry?

y = x

→ x = -42⋅1

= -42

x = 2

^{2}+ 4x - 9→ x = -42⋅1

= -42

x = 2

Close

### Example

y = -5x

Axis of symmetry?

Solution ^{2}+ xAxis of symmetry?

y = -5x

→ x = -12⋅(-5)

x = 110

^{2}+ x→ x = -12⋅(-5)

x = 110

Close

## Quadratic Function: Vertex

### Formula

y = a(x - h)

The vertex (red point)^{2}+ kis the lowest point of the quadratic function.

(highest point: when a is (-).)

To find the vertex of a quadratic function,

change the quadratic function

to vertex form.

y = a(x - h)

^{2}+ k

Then the vertex is (h, k).

The vertex is on the axis of symmetry.

So x = h is the axis of symmetry.

### Example

y = x

Vertex?

Solution ^{2}- 4x + 5Vertex?

y = x

= x

= (x - 2)

= (x - 2)

(2, 1)

^{2}- 4x + 5= x

^{2}- 2⋅x⋅2 + 2^{2}- 2^{2}+ 5 - [1]= (x - 2)

^{2}- 4 + 5 - [2]= (x - 2)

^{2}+ 1(2, 1)

[1]

Use x

to make a perfect square trinomial.

x

→ x

+2

So, to undo the change, write -2

And write +5.

Completing the Square

^{2}- 4xto make a perfect square trinomial.

x

^{2}- 4x→ x

^{2}- 2⋅x⋅2 + 2^{2}+2

^{2}is added.So, to undo the change, write -2

^{2}.And write +5.

Completing the Square

Close

### Example

y = x

Vertex?

Solution ^{2}+ 6x - 1Vertex?

y = x

= x

= (x + 3)

= (x - (-3))

(-3, -10)

^{2}+ 6x - 1= x

^{2}+ 2⋅x⋅3 + 3^{2}- 3^{2}- 1= (x + 3)

^{2}- 9 - 1= (x - (-3))

^{2}- 10(-3, -10)

Close

### Example

y = -x

Vertex?

Solution ^{2}+ 8x - 16Vertex?

y = -x

= -(x

= -(x

= -(x - 4)

= -(x - 4)

(4, 0)

^{2}+ 8x - 16= -(x

^{2}- 8x) - 16= -(x

^{2}- 2⋅x⋅4 + 4^{2}) + 4^{2}- 16 - [1]= -(x - 4)

^{2}+ 16 - 16= -(x - 4)

^{2}(4, 0)

[1]

Use x

to make a perfect square trinomial.

-(x

→ -(x

-(+4

So, to undo the change, write +4

And write -16.

^{2}- 8xto make a perfect square trinomial.

-(x

^{2}- 8x)→ -(x

^{2}- 2⋅x⋅4 + 4^{2})-(+4

^{2}) is added.So, to undo the change, write +4

^{2}.And write -16.

Close

## Quadratic Function: Finding Zeros

### Formula

y = a(x - r

The zeros is the intersection_{1})(x - r_{2})of the graph and the x-axis.

So, to find the zeros of the quadratic function,

1. Factor the quadratic function.

2. Set (left side) = 0.

3. Solve the quadratic equation: (left side) = 0.

### Example

y = x

Zeros?

Solution ^{2}- 2x - 3Zeros?

y = x

= (x + 1)(x - 3) = 0 - [1] [2]

1) x + 1 = 0

x = -1

2) x - 3 = 0

x = 3

x = -1, 3

^{2}- 2x - 3= (x + 1)(x - 3) = 0 - [1] [2]

1) x + 1 = 0

x = -1

2) x - 3 = 0

x = 3

x = -1, 3

Graph

[2]

To find the zeros,

set (left side) = 0

and solve (x + 1)(x - 3) = 0.

set (left side) = 0

and solve (x + 1)(x - 3) = 0.

Close

### Example

y = -3x

Zeros?

Solution ^{2}+ 12Zeros?

y = -3x

= -3(x

= -3(x

= -3(x + 2)(x - 2) = 0

1) x + 2 = 0

x = -2

2) x - 2 = 0

x = 2

x = -2, 2

^{2}+ 12= -3(x

^{2}- 4)= -3(x

^{2}- 2^{2})= -3(x + 2)(x - 2) = 0

1) x + 2 = 0

x = -2

2) x - 2 = 0

x = 2

x = -2, 2

Graph

Close

## Quadratic Function: Number of Zeros

### Formula

y = ax

→ D = b

D > 0D = 0D < 0

Just like the D of a quadratic equation,^{2}+ bx + c→ D = b

^{2}- 4acD > 0D = 0D < 0

you can find the number of zeros

by using the D of the quadratic function.

(without solving the quadratic function)

### Example

y = x

Solution ^{2}+ 8x - 3 y = x

D = 8

= 64 + 12

= 76 > 0

Two zeros

^{2}+ 8x - 3D = 8

^{2}- 4⋅1⋅(-3)= 64 + 12

= 76 > 0

Two zeros

Close

### Example

y = -4x

Solution ^{2}+ 4x - 1 y = -4x

D = 4

= 16 - 16

= 0

One zero

^{2}+ 4x - 1D = 4

^{2}- 4⋅(-4)⋅(-1)= 16 - 16

= 0

One zero

Close

### Example

y = 2x

Solution ^{2}- x + 7 y = 2x

D = (-1)

= 1 - 56

= -55 < 0

No zeros

^{2}- x + 7D = (-1)

^{2}- 4⋅2⋅7= 1 - 56

= -55 < 0

No zeros

Close

## Quadratic Inequality

### Example

x

Solution ^{2}- 3x - 10 ≤ 0 x

(x + 2)(x - 5) ≤ 0 - [1]

1) x + 2 = 0

x = -2

2) x - 5 = 0

x = 5

x = -2, 5 - [2]

- [3] [4]

-2 ≤ x ≤ 5

^{2}- 3x - 10 ≤ 0(x + 2)(x - 5) ≤ 0 - [1]

1) x + 2 = 0

x = -2

2) x - 5 = 0

x = 5

x = -2, 5 - [2]

-2 ≤ x ≤ 5

[2]

The zeros are -2 and 5.

[3]

Draw y = (x + 2)(x - 5) on the x-axis.

[4]

x

The left side is less than (or equal to) 0.

So color the region

where the graph is below the x-axis (y = 0).

The inequality sign, ≤, does include '='.

So draw full circles on the zeros:

-2 and 5.

^{2}- 3x - 10 ≤ 0The left side is less than (or equal to) 0.

So color the region

where the graph is below the x-axis (y = 0).

The inequality sign, ≤, does include '='.

So draw full circles on the zeros:

-2 and 5.

Close

### Example

x

Solution ^{2}- 16 > 0 x

x

(x + 4)(x - 4) > 0

1) x + 4 = 0

x = -4

2) x - 4 = 0

x = 4

x = -4, 4 - [1]

- [2] [3]

x < -4 or x > 4

^{2}- 16 > 0x

^{2}- 4^{2}> 0(x + 4)(x - 4) > 0

1) x + 4 = 0

x = -4

2) x - 4 = 0

x = 4

x = -4, 4 - [1]

x < -4 or x > 4

[1]

The zeros are -4 and 4.

[2]

Draw y = (x + 4)(x - 4) on the x-axis.

[3]

x

The left side is greater than 0.

So color the region

where the graph is above the x-axis (y = 0).

The inequality sign, >, does not include '='.

So draw empty circles on the zeros:

-4 and 4.

^{2}- 4^{2}> 0The left side is greater than 0.

So color the region

where the graph is above the x-axis (y = 0).

The inequality sign, >, does not include '='.

So draw empty circles on the zeros:

-4 and 4.

Close

### Example

-x

Solution ^{2}+ 10x - 25 ≥ 0 -x

x

x

(x - 5)

x - 5 = 0

x = 5 - [1]

- [2] [3]

x = 5

^{2}+ 10x - 25 ≥ 0x

^{2}- 10x + 25 ≤ 0x

^{2}- 2⋅x⋅5 + 5^{2}≤ 0(x - 5)

^{2}≤ 0x - 5 = 0

x = 5 - [1]

x = 5

[1]

The zero is 5.

[2]

Draw y = (x - 5)

^{2}on the x-axis.[3]

(x - 5)

The left side is less than (or equal to) 0.

So color the region

where the graph is below the x-axis (y = 0).

(no region to color)

The inequality sign does include '='.

So draw a full circle on the zero:

x = 5.

^{2}≤ 0The left side is less than (or equal to) 0.

So color the region

where the graph is below the x-axis (y = 0).

(no region to color)

The inequality sign does include '='.

So draw a full circle on the zero:

x = 5.

Close

## System of Equations: Quadratic-Linear

### Example

y = x

y = x + 4

Solution ^{2}- 2xy = x + 4

y = x

y = x + 4

x

x

(x + 1)(x - 4) = 0 - [2]

1) x + 1 = 0

x = -1

y = -1 + 4 - [3]

= 3

→ (-1, 3)

2) x - 4 = 0

x = 4

y = 4 + 4 - [4]

= 8

→ (4, 8)

(-1, 3), (4, 8)

^{2}- 2xy = x + 4

x

^{2}- 2x = x + 4 - [1]x

^{2}- 3x - 4 = 0(x + 1)(x - 4) = 0 - [2]

1) x + 1 = 0

x = -1

y = -1 + 4 - [3]

= 3

→ (-1, 3)

2) x - 4 = 0

x = 4

y = 4 + 4 - [4]

= 8

→ (4, 8)

(-1, 3), (4, 8)

Graph

[1]

The right sides are equal. (y)

So x

So x

^{2}- 2x = x + 4.[3]

[4]

x = 4 → y = x + 4

Close

### Number of Intersecting Points

D > 0D = 0D < 0

you'll get a quadratic equation.

The D of the quadratic equation determines

the number of the intersecting points.

D > 0: 2 intersecting points

D = 0: 1 intersecting point

D < 0: no intersecting points

### Example

Find the range of k that makes the given functions intersect.

y = x

y = x + k

Solution y = x

^{2}- xy = x + k

y = x

y = x + k

x

x

D = (-2)

4 + 4k ≥ 0

4k ≥ -4

k ≥ -1

^{2}- xy = x + k

x

^{2}- x = x + kx

^{2}- 2x - k = 0D = (-2)

^{2}- 4⋅1⋅(-k)4 + 4k ≥ 0

4k ≥ -4

k ≥ -1

[1]

To make the given functions intersect,

there should be

at least one intersecting point:

one intersecting point (D = 0)

or

two intersecting points (D > 0).

So set D ≥ 0.

Discriminant

there should be

at least one intersecting point:

one intersecting point (D = 0)

or

two intersecting points (D > 0).

So set D ≥ 0.

Discriminant

Close