Preparing for this section

- Intercepts
- Completing the Square
- Quadratic Equations
- Graphing Techniques

Objectives

1. Finding Vertex and Axis of Symmetry

2. Graphing Quadratic Functions

3. Using Desmos to Find the Quadratic Function of Best Fit to Data

A quadratic function is one of the form

, whereƒ(x) = ax^{2}+ bx + c,a, andbare numbers withcnot equal to zero.a

Quadratic functions are widely used in science, business and engineering. The U-shape of a parabola can describe the trajectories of a bouncing ball and of water jets in a fountain. Most of the objects we use every day, from cars to clocks, would not exist if someone, somewhere hadn't applied quadratic functions to their design.

Here’s an example: you dive from a platform 3 meters high into the water. Let’s say you hit the water 4 meters away from the platform. That’s how your path is going to look like:

##### 1. Finding Vertex and Axis of Symmetry

$ƒ(x)=ax^2+bx+c, a≠0$

$Vertex=( -\dfrac{b}{2a},ƒ(- \dfrac{b}{2a}))$

Parabola opens up if a>0; vertex is a minimum point.

Parabola opens down if a<0; vertex is a maximum point.

Axis of symmetry: $x = -\dfrac{b}{2a}$

Example 1

Find the coordinates of the vertex of $ƒ\left(x\right)=4x^2-3x+2$. Without graphing, determine if the vertex is the maximum or minimum point of the quadratic function.

Solution

This quadratic function is written in the form $ƒ(x)=ax^2+bx+c, a≠0$. That means we can find the vertex using the formula $Vertex=( -\dfrac{b}{2a},ƒ(- \dfrac{b}{2a}))$.

$a=4,b=-3,c=2$

$x=- \dfrac{b}{2a}=- \dfrac{-3}{2(4)}=\dfrac{3}{8}=0.375$

$ƒ(-\dfrac{b}{2a})=ƒ(\dfrac{3}{8})=0.5625-1.125+2=1.4375$

The Vertex would be $(0.375;1.4375)$.

Axis of Symmetry:

$x = -\dfrac{b}{2a}$

$x = -\dfrac{-3}{2(4)}$

$x = 0.375$

Next we need to determine if the vertex that we found is a maximum or a minimum point, without graphing. Knowing which direction the curve opens can help us answer this question. Since a=4, and 4 is more than 0, this parabola would open up. That means that the vertex is a minimum point.

##### 2. Graphing Quadratic Functions

From the information we got in Example 1 we can graph $ƒ(x)=ax^2+bx+c, a≠0$, by hand.

To graph a quadratic function we need to follow those 4 steps

**Does the graph open up or down?****Find the Vertex****Find the Intercepts****Graph the Parabola**

Example 2

Use the vertex and the intercepts to graph the function. Find the equation for this function's axis of symmetry.

Solution

**Step 1. Does the Graph Open Up or Down?**

a = 4 and 4 > 0, so as we already found out that this parabola opens up.

**Step 2. Find the Vertex**

To find the vertex we can use the formula: $Vertex=( -\dfrac{b}{2a},ƒ(- \dfrac{b}{2a}))$

But since we know that our vertex is $(0.375;1.4375)$, we can move on to the next step.

**Step 3. Find the Intercepts**

y-intercept

The y-intercept is always where the graph crosses the y-axis, which means x=0.

$ƒ\left(x\right)=4x^2-3x+2$

$ƒ\left(0\right)=4(0)^2-3(0)+2$

The y-intercept is $(0;2)$

x-intercept

The x-intercept is always where the graph crosses the x-axis, which means y=0

$ƒ\left(x\right)=4x^2-3x+2$

$0=4x^2-3x+2$

This does not factor, so let's try using the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-4\pm\sqrt{(-3)^2-4(4)(2)}}{2(4)}$

$x=\frac{-4\pm\sqrt{-23}}{8}$

The number under the square root is negative so there is no real number solution. Also there are NO x-intercepts.

**Step 4. Graph the parabola**

$ƒ\left(x\right)=4x^2-3x+2$

y-int=(0;2)

x-int=none

vertex:(0.375;1.4375)

axis of symmetry: x=0.375

a = 4 > 0, opens up

##### 3. Using Desmos to Find the Quadratic Function of Best Fit to Data

Example 3

The marketer sells around 100 bananas a day. The cost of one banana is $0.80. The marketer counted that for every $0.05 price increase 2 fewer bananas will be sold. Graph and determine what cost will maximize the profit?

Solution

Let's say x is the number of 0.05 price increases.

Then 0.8+0.05x is the single price,

so 100-2x is the number of bananas sold.

Now, to find the income we need to multiply the number of bananas sold by the price.

$Income=(100-2x)(0.8+0.05x)$

$Income=-0.1x^2+3.4x+80$

We got a quadratic function. a=-0.1, which means that the parabola opens down, so x is the number of prices increases needed to maximize the profit.

To find x we will use the formula:

$x = -\dfrac{b}{2a}$

$x = -\dfrac{3.4}{2(-0.1)}$

$x = 17$

This means that the profit will be maximized after 17 $0.05 price increases.

So one banana will cost:

$Banana = 0.8+0.05(17)$

$Banana = 1.65$

The most profitable price of one banana is $1.65.

Now we can find the maximum income.

$Max.Income = (100-2x)(0.8+0.05x)$

$Max.Income = (100-34)(0.8+0.85)$

$Max.Income = (66)(1.65)$

$Max.Income = 108.9$

The answer is $108.9

## Skill building

- For each of the following quadratic functions find the vertex, axis of symmetry, whether the parabola opens up or down and sketch it:
- $y = x^2-2x+2$
- $y = 2x^2-x+6$
- $y = 4x^2-2x+8$
- $y = -4x^2-4x+1$
- $y = 0.2x^2-0.1x+0.0125$
- $y = x(x+5)-10$

- Using quadratic formula, find solutions to each of the following equations
- $x^2-4x+2 = 0$
- $x^2 = 10x-15$
- $x(x+5) = -10$
- $-x^2+31x-50 = 0$

## Answers

- The answers are in order: vertex, axis of symmetry, up(down).
- (1;1); 1; up.
- (0.25;5.875); 0.25; up.
- (0.25;7.75); 0.25; up.
- (-0.5;2); -0.5; down.
- (0.25;0); 0.25; up.
- (2.5;8.75); 2.5; up.

- Rounded to hundredth
- 0.58 and 3.41
- 1.84 and 8.16
- no solution
- 29.29 and 1.7