Learn How to Graph a Quadratic Function

Name	Equation	Characteristics
Standard Form	$y = a x^{2} + b x + c$	$a,$ $b,$ and $c$ are real numbers, $a \neq = 0,$ and $c$ is the $y -$ intercept of the parabola.
Vertex Form	$y = a (x - h)^{2} + k$	$a,$ $h,$ and $k$ are real numbers, $a \neq = 0,$ and $(h, k)$ is the vertex of the parabola.
Intercept Form (also called Factored Form)	$y = a (x - x_{1}) (x - x_{2})$	$a,$ $x_{1},$ and $x_{2}$ are real numbers, $a \neq = 0,$ and $x_{1}$ and $x_{2}$ are the $x -$ intercepts of the parabola.

Name

Equation

Characteristics

y = a x^{2} + b x + c

a,

b,

and

c

are real numbers,

a \neq = 0,

and

c

is the

y -

intercept of the parabola.

Vertex Form

y = a (x - h)^{2} + k

a,

h,

and

k

are real numbers,

a \neq = 0,

and

(h, k)

is the vertex of the parabola.

Intercept Form
(also called Factored Form)

y = a (x - x_{1}) (x - x_{2})

a,

x_{1},

and

x_{2}

are real numbers,

a \neq = 0,

and

x_{1}

and

x_{2}

are the

x -

intercepts of the parabola.

Direction	Vertex	Axis of Symmetry
$a = - \frac{1}{4}$	$h = 4$ and $k = 8$	$h = 4$
Since $- \frac{1}{4}$ is less than $0,$ the parabola opens downward.	The vertex is located at $(4, 8) .$	The axis of symmetry is the vertical line $x = 4 .$

Direction

Vertex

Axis of Symmetry

a = - \frac{1}{4}

h = 4

and

k = 8

h = 4

Since

- \frac{1}{4}

is less than

0,

the parabola opens downward.

The vertex is located at

(4, 8) .

The axis of symmetry is the vertical line

x = 4 .

Function $1$	Function $2$	Function $3$
$y = 2 (x + 1)^{2} + 7$ $⇕$ $y = 2 (x - (- 1))^{2} + 7$	$y = (x - 3)^{2} + 1$ $⇕$ $y = 1 (x - 3)^{2} + 1$	$y = 5 (x - 2)^{2} - 3$ $⇕$ $y = 5 (x - 2)^{2} + (- 3)$

Function

1

Function

2

Function

3

y = 2 (x + 1)^{2} + 7

⇕

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

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

⇕

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

y = 5 (x - 2)^{2} - 3

⇕

y = 5 (x - 2)^{2} + (- 3)

$y = a (x - h)^{2} + k$
Direction	Vertex	Axis of Symmetry
$a > 0 a < 0 \Rightarrow upward \Rightarrow downward$	$(h, k)$	$x = h$

y = a (x - h)^{2} + k

Direction

Vertex

Axis of Symmetry

a > 0 a < 0 \Rightarrow upward \Rightarrow downward

(h, k)

x = h

To identify the vertex of the parabola, the given equation can be compared to the general form of a quadratic function written in vertex form.

General Form y = a (x - h)^{2} + k Given Function y = - 2 (x - \frac{3}{2})^{2} + \frac{9}{2}

For the given function,

h = \frac{3}{2}

and

k = \frac{9}{2} .

This means that the vertex of the parabola is

(\frac{3}{2}, \frac{9}{2}) .

Since the parabola illustrates the position of the ball, the first quadrant will only be considered.

The axis of symmetry of the parabola is the vertical line through the vertex. Therefore, in this case, the equation of the axis of symmetry is $x = \frac{3}{2} .$

The

y -

intercept will now be determined and plotted. To do so,

x = 0

will be substituted in the given equation.

y = - 2 (x - \frac{3}{2})^{2} + \frac{9}{2}

Substitute

$x = 0$

y = - 2 (0 - \frac{3}{2})^{2} + \frac{9}{2}

Evaluate right-hand side

SubTerm

Subtract term

y = - 2 (- \frac{3}{2})^{2} + \frac{9}{2}

NegBaseToPosPow

$(- a)^{2} = a^{2}$

y = - 2 (\frac{3}{2})^{2} + \frac{9}{2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

y = - 2 (\frac{9}{4}) + \frac{9}{2}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

y = - \frac{1 8}{4} + \frac{9}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

y = - \frac{9}{2} + \frac{9}{2}

AddFrac

Add fractions

y = 0

The parabola intercepts the

y -

axis at

(0, 0) .

Next, the

y -

intercept will be reflected across the axis of symmetry.

Note that for the given function, the value of

a

- 2 .

Therefore, the parabola opens downward. This corresponds to the loci of the points plotted in the coordinate plane. Finally, these points can be connected with a smooth curve to draw the parabola.

Direction	Zeros	Axis of Symmetry
$a = \frac{1}{2}$	$p = 7$ and $q = 13$	$\frac{p + q}{2}$ $⇓$ $\frac{7 + 1 3}{2} = 10$
Since $\frac{1}{2}$ is greater than $0,$ the parabola opens upward.	The zeros are $7$ and $13 .$ Therefore, the parabola intersects the $x -$ axis at $(7, 0)$ and $(13, 0) .$	The axis of symmetry is the vertical line $x = 10 .$

Direction

Zeros

Axis of Symmetry

a = \frac{1}{2}

p = 7

and

q = 13

\frac{p + q}{2}

⇓

\frac{7 + 1 3}{2} = 10

Since

\frac{1}{2}

is greater than

0,

the parabola opens upward.

The zeros are

7

and

13 .

Therefore, the parabola intersects the

x -

axis at

(7, 0)

and

(13, 0) .

The axis of symmetry is the vertical line

x = 10 .

Function $1$	Function $2$	Function $3$
$y = 2 (x + 1) (x - 3)$ $⇕$ $y = 2 (x - (- 1)) (x - 3)$	$y = (x - 5) (x - 9)$ $⇕$ $y = 1 (x - 5) (x - 9)$	$y = 5 x (x - 2)$ $⇕$ $y = 5 (x - 0) (x - 2)$

Function

1

Function

2

Function

3

y = 2 (x + 1) (x - 3)

⇕

y = 2 (x - (- 1)) (x - 3)

y = (x - 5) (x - 9)

⇕

y = 1 (x - 5) (x - 9)

y = 5 x (x - 2)

⇕

y = 5 (x - 0) (x - 2)

$y = a (x - p) (x - q)$
Direction	Zeros	Axis of Symmetry
$a > 0 a < 0 \Rightarrow upward \Rightarrow downward$	$p$ and $q$	$x = \frac{p + q}{2}$

y = a (x - p) (x - q)

Direction

Zeros

Axis of Symmetry

a > 0 a < 0 \Rightarrow upward \Rightarrow downward

p

and

q

x = \frac{p + q}{2}

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Catch-Up and Review

y-intercept

Example

Extra

Answer

Hint

Solution

Hint

Solution

Example

Extra

Answer

Hint

Solution

Hint

Solution

$y -$ intercept