2. Standard Form of a Quadratic Function - 4. Quadratic Functions and Equations

We want to draw the graph of a quadratic function written in standard form.

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

To do so, we will follow five steps.

Identify $a,$ $b,$ and $c .$
Calculate and sketch the axis of symmetry.
Find and plot the vertex.
Find and plot the $y -$ intercept and its symmetric point across the axis of symmetry.
Draw a smooth curve through the three plotted points.

Let's do it!

Identify $a,$ $b,$ and $c$

We will start by identifying the values of

a,

b,

and

c .

y = x^{2} - 2 x + 4 ⇕ y = 1 x^{2} + (- 2) x + 4

We have identified that

a = 1,

b = - 2,

and

c = 4 .

Axis of Symmetry

The axis of symmetry is the vertical line that divides the parabola into two mirror images. Its equation follows a specific formula.

x = - \frac{b}{2 a}

Let's substitute our given values

a = 1

and

b = - 2

into this equation.

x = - \frac{b}{2 a}

SubstituteII

$a = 1$ , $b = - 2$

x = - \frac{- 2}{2 ( 1 )}

▼

Simplify right-hand side

MultByOne

$a \cdot 1 = a$

x = - \frac{- 2}{2}

MoveNegFracToNum

Put minus sign in numerator

x = \frac{- ( - 2 )}{2}

NegNeg

$- (- a) = a$

x = \frac{2}{2}

QuotOne

$\frac{a}{a} = 1$

x = 1

The axis of symmetry is the line

x = 1 .

Vertex

To find the vertex of the parabola, we will need to think of

y

as a function of

x,

y = f (x) .

We can write the expression for the vertex by stating the

x -

and

y -

coordinates in terms of

a

and

b .

Vertex : (- \frac{b}{2 a}, f (- \frac{b}{2 a}))

When determining the axis of symmetry, we found that

- \frac{b}{2 a} = 1 .

Therefore, the

x -

coordinate of the vertex is

1

and the

y -

coordinate is

f (1) .

To find this value, substitute our

x -

coordinate for

x

in the given equation.

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

Substitute

$x = 1$

y = 1^{2} - 2 (1) + 4

▼

Simplify right-hand side

CalcPow

Calculate power

y = 1 - 2 (1) + 4

MultByOne

$a \cdot 1 = a$

y = 1 - 2 + 4

AddSubTerms

Add and subtract terms

y = 3

The vertex of the parabola is

(1, 3) .

$y -$ Intercept and Symmetric Point

From our equation we know that $c = 4,$ so the $y -$ intercept is $4 .$ Let's plot this point and the one across the axis of symmetry.

Graph

Since $a = 1,$ which is greater than zero, we can confirm that our parabola opens upwards. Let's draw a smooth curve connecting the three points we have. We should not use a straight edge for this!

Extra

A Common Mistake

One common mistake when identifying the key features of a parabola algebraically is forgetting to include the negatives in the values of these constants. The standard form is addition only, so any subtraction must be treated as negative values of

a,

b,

c .

Let's look at an example.

\underline{a x^{2} + b x + c} y = 3 x^{2} - 4 x - 2 ⇕ y = 3 x^{2} + (- 4 x) + (- 2)

In this case, the values of

a,

b,

and

c

are

3,

- 4,

and

- 2 .

They are not

3,

4,

and

2 .

a = 3, b = 4, c = 2 \times a = 3, b = - 4, c = - 2 ✓

Exercise 2 Page 206

Hints

Check the answer

Practice exercises

Asses your skill

Identify $a,$ $b,$ and $c$

Axis of Symmetry

Vertex

$y -$ Intercept and Symmetric Point

Graph

Extra

Exercise 2 Page 206

Hints

Check the answer

Practice exercises

Asses your skill

Identify a, b, and c

Axis of Symmetry

Vertex

y-Intercept and Symmetric Point

Graph

Extra

Identify $a,$ $b,$ and $c$

$y -$ Intercept and Symmetric Point