Unit 3.1 – Quadratic Functions & Models

Polynomial functions are classified by degree. For instance, a constant

function f (x) = c with c 0 has degree 0, and a linear function f (x) = ax + b

with a 0 has degree 1. In this section, you will study second-degree

polynomial functions, which are called Quadratic Functions.

For instance, each of the following functions is a quadratic function.

f(x) = –2x2 – 2x – 2 Standard Form (ax2 + bx + c, where a ≠ 0)

g(x) = 2(x + 1)2 – 3 Vertex Form (useful for transformations)

m(x) = 2(x – 1)(x + 1) Factored Form (useful for x-intercepts)

Know the book and the homework call the second one Standard Form.

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The graph of a quadratic function is a special type of “U”-shaped curve called a

Parabola. Parabolas occur in many real-life applications – especially those involving

reflective properties of satellite dishes and flashlight reflectors.

All parabolas are symmetric with respect to a line called the Axis of Symmetry, or

simply the Axis of the parabola. The point where the axis intersects the parabola is the

Vertex of the parabola.

(Minimum)

(Maximum)

Leading coefficient is positive. Leading coefficient is negative.

The Vertex Form of a quadratic function is f(x) = a(x – h)2 + k where a ≠ 0

This form is especially convenient for sketching a parabola because it identifies the

vertex of the parabola as (h, k). Plus, it identifies four basic transformations of the graph:

Left shift of two units followed by

a downward shift of three units

How do you graph a quadratic function in Standard Form? One way is to create a table of

x and y values, then plot them individually. Another way is to convert it to Vertex Form,

and use the h and k values to locate the vertex (and by default, the axis). You will use

the concept of Completing The Square (CTS).

Sketch the graph of f(x) = 2x2 + 8x + 7. Identify the vertex and the axis of the parabola.

f(x) = 2(x2 + 4x) + 7 1st step in CTS: Factor out the x2 coefficient that’s not 1.

BUT: Limit it to x2 and x terms.

2nd step in CTS: Calculate (𝑏

2)2

f(x) = 2(x2 + 4x + 4 – 4) + 7 3rd step in CTS: Add it within the parenthesis.

BUT: Immediately subtract it.

f(x) = 2(x2 + 4x + 4) – 8 + 7 Remove that subtraction via distribution.

f(x) = 2(x + 2)(x + 2) – 8 + 7 Factor the parenthesis.

f(x) = 2(x + 2)2 – 1 Simplify.

f(x) = 2(x + 2)2 – 1 h = –2 k = –1

The graph opens upwards.

The vertex is at (–2,–1), so the axis is at –2.

This corresponds to a left shift of two units

and a downward shift of one unit, relative

to the graph of y = 2x2.

As a reminder, to find the y-intercept, set x to

zero: 2(x + 2)2 – 1 = 2(0 + 2)2 – 1 = 8 – 1 = 7

It was clearly visible in its Standard Form:

f(x) = 2x2 + 8x + 7

To find the x-intercepts (roots) in this case, use

the Quadratic Formula. They’re −2 ±2

2

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Sketch the graph of f(x) = 3x2 + 6x + 4. Identify the vertex and the axis of the parabola.

f(x) = 3(x2 + 2x) + 4 1st step in CTS: Factor out the x2 coefficient that’s not 1.

BUT: Limit it to x2 and x terms.

2nd step in CTS: Calculate (𝑏

2)2

f(x) = 3(x2 + 2x + 1 – 1) + 4 3rd step in CTS: Add it within the parenthesis.

BUT: Immediately subtract it.

f(x) = 3(x2 + 2x + 1) – 3 + 4 Remove that subtraction via distribution.

f(x) = 3(x + 1)(x + 1) – 3 + 4 Factor the parenthesis.

f(x) = 3(x + 1)2 + 1 Simplify.

The graph opens upwards. The vertex is at (–1,1),

so the axis is at –1. And the y-intercept is 4.

To find x-intercepts (zeros), factor and set each to zero.

In this case, it’s not factorable. (Discriminant is less than 0.)

CW # 1

f(x) = –2x2 – 2x – 2 Standard Formg(x) = 2(x + 1)2 – 3 Vertex Form m(x) = 2(x – 1)(x + 1) Factored Form

CW # 2Convert these functions. f(x) = –2x2 – 2x – 2 g(x) = 2(x + 1)2 – 3m(x) = 2(x – 1)(x + 1)

Write the standard form of the quadratic function whose graph is a parabola

with vertex (1,2) and that passes through the point (3,–6).

The vertex is (h,k) = (1,2), so the equation has the form:

All you need is the value of a.

The parabola passes through the point (3,–6), so it follows that f(3) = –6. So:

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CW # 3

We had given you a quadratic equation in Standard Form, and asked you to find the

vertex (and by default, the axis, which is its x-portion). And until now you converted the

equation into Vertex Form first. But you can find the vertex from Standard Form itself,

then use that to find the vertex.

The axis = 𝒙 =−𝒃

𝟐𝒂

Once you have that x value, substitute it into the equation to find the corresponding y

value; together, they ARE the vertex.

Find the vertex for the ones we’ve already done.

f(x) = 2x2 + 8x + 7

f(x) = 3x2 + 6x + 4

f(x) = 3x2 – 6x + 4

CW # 4

Sam plays on the school soccer team. The path of Sam’s ball when kicked towards the

goal can be represented by the quadratic function y = −0.18x2 + 1.48x where x is the

horizontal distance (in feet) and y is the height (in feet).

How far out is the ball when it reaches its maximum height? That’s the x coordinate of

the vertex, aka axis.

𝑥 =−𝑏

2𝑎=

−1.48

2(−0.18)= 4.11 feet

What’s the maximum height of

the ball? That’s the y coordinate

of the vertex. Substitute the

x coordinate into the equation.

y = −0.18(4.11)2 + 1.48(4.11) =

= 3.04 feet

How far did he kick the ball? That’s the x-intercept, aka Root/Zero/Solution. Because you

can’t factor it, or use square roots, or complete the square - use the Quadratic Formula

to find it: 8.22 feet. In this case, you could just double the axis: 4.11 · 2 = 8.22

An athlete whose event is the shot put releases a shot. When the shot

whose path is shown by the graph is released at an angle of 55°, its

height in feet can be modeled by 𝑓 𝑥 = −0.02𝑥2 + 1.4𝑥 + 5.8, where x

is the shot's horizontal distance, in feet, from its point of release.

What is the maximum height

of the shot? ________

(y coordinate of vertex)

How far from its point of release

does this occur? ________

(x coordinate of vertex)

From what height was the shot

released? ________ (y-intercept)

What is the shot's maximum horizontal

distance? ________ (x-intercept)

CW # 5