Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

As you can see the St. Louis Arch is an example of a parabola in real life. This parabola would be a negative parabola because it is facing down. Parabolas are used very often but people don’t tend to notice. Another example would be the McDonald’s golden arches.

But every parabola is different. So how do we know the equation of the parabola we want?

This lesson focuses on the connection between the equations of quadratics and their graphs, and how we can use the idea of transformations to more easily identify and make sense of these quadratics.

The word parabola may not be a common word in everyday life, but the concept of parabolas is most definitely not uncommon. From bridges to radio telescopes, they are not hard to spot. In fact, many famous landmarks are parabolas e.g. the Sydney Harbour Bridge. Parabolas also appear as the trajectory that objects follow once thrown. For example, when you try to make a basketball shot it follows the path of a parabola.

Given how common parabolas are, it would be useful to graph them and know their equations. For example, when avalanche control experts want to trigger an avalanche, they need to know where the detonators will land on the mountain side. If the detonator follows a parabolic path, its equation can help determine such unknowns.

All parabolas can be thought of as some transformation of the standard parabola which has the equation $y=x^2$`y`=`x`2.

What would be the equation of a parabola that is flatter or narrower and which has a vertex that is not at $\left(0,0\right)$(0,0)?

Concavity is a word used to describe the shape of a curve.

Curves can be concave up (making a cup like shape, or a smiling mouth,)

Curves can be concave down (making a hill shape or a sad mouth, )

$y=x^2$`y`=`x`2

- Notice that as the $x$
`x`values we substitute get larger, i.e. $x=2$`x`=2, $x=5$`x`=5, $x=50$`x`=50 etc, $y$`y`increases towards infinity. - Notice that as the $x$
`x`values we substitute get negatively larger, i.e. $x=-2$`x`=−2, $x=-5$`x`=−5, $x=-50$`x`=−50 etc, $y$`y`still increases towards infinity. - $y$
`y`can never have a negative value. - This parabola is
**concave up**.

$y=-x^2$`y`=−`x`2

- Notice that as the $x$
`x`values we substitute get larger, i.e. $x=2$`x`=2, $x=5$`x`=5, $x=50$`x`=50 etc, $y$`y`decreases towards negative infinity. - Notice that as the $x$
`x`values we substitute get negatively larger, i.e. $x=-2$`x`=−2, $x=-5$`x`=−5, $x=-50$`x`=−50 etc, $y$`y`still decreases towards negative infinity. - $y$
`y`can never have a positive value. - This parabola is
**concave down**.

The concavity of a parabola is determined by the $a$`a` value in the equation $y=ax^2$`y`=`a``x`2.

- $y=-x^2$
`y`=−`x`2 vs $y=x^2$`y`=`x`2: The sign of the coefficient $a$`a`determines whether the parabola is concave up or concave down.

Compare the parabolas $y=\frac{1}{2}x^2$`y`=12`x`2, $y=x^2$`y`=`x`2, and $y=2x^2$`y`=2`x`2. The larger the numerical value of $a$`a`, the more quickly $y$`y` increases, i.e. the narrower the parabola. In this case $y=2x^2$`y`=2`x`2 will be the narrowest.

So just by looking at the coefficient of $x^2$`x`2, we can determine some features of the parabola's shape:

- $y=\frac{1}{2}x^2$
`y`=12`x`2: concave up, wider than $y=x^2$`y`=`x`2. - $y=-x^2$
`y`=−`x`2: concave down, same narrowness as $y=x^2$`y`=`x`2. - $y=-3x^2$
`y`=−3`x`2: concave down, narrower than $y=-x^2$`y`=−`x`2, the result of reflecting $y=3x^2$`y`=3`x`2 about the $x$`x`-axis.

Every parabola has a vertex. If it is concave up, the vertex is the point where the $y$`y` value is a minimum. If it is concave down, the vertex is the point where the $y$`y` value is a maximum. Keeping this in mind, let's consider the minimum and maximum value of each parabola below to locate its vertex.

**1)** $y=x^2$`y`=`x`2 (concave up so $y$`y` has a minimum value at vertex)

- Minimum value of $y$
`y`: $y$`y`is the result of squaring a value so the smallest possible value is $y=0$`y`=0 - $x$
`x`value corresponding to minimum $y$`y`value: we would need to substitute $x=0$`x`=0 to get $y=0$`y`=0 - Vertex: $\left(0,0\right)$(0,0)

**2) **$y=x^2+2$`y`=`x`2+2 (same concavity as $y=x^2$`y`=`x`2)

- Minimum value of $y$
`y`: $y=x^2$`y`=`x`2 has a minimum value of $0$0, so adding $2$2 to it, the smallest possible value is $y=2$`y`=2 - $x$
`x`value corresponding to minimum $y$`y`value: we would need to substitute $x=0$`x`=0 to get $y=2$`y`=2 - Vertex: $\left(0,2\right)$(0,2)

$y=x^2+2$`y`=`x`2+2 is the result of a translation of $y=x^2$`y`=`x`2 by $2$2 units upwards.

**3)** $y=-x^2-5$`y`=−`x`2−5 (same concavity as $y=-x^2$`y`=−`x`2)

- Maximum value of $y$
`y`: $y=-x^2$`y`=−`x`2 has a maximum value of $0$0, so subtracting $5$5 from it, the maximum value of $y$`y`here is $y=-5$`y`=−5 - $x$
`x`value corresponding to minimum $y$`y`value: we would need to substitute $x=0$`x`=0 to get $y=-5$`y`=−5 - Vertex: $\left(0,-5\right)$(0,−5)

$y=-x^2-5$`y`=−`x`2−5 is the result of a translation of $y=-x^2$`y`=−`x`2 by $5$5 units downwards.

Considering vertical translations only, the equations we encounter will be of the form $y=ax^2+k$`y`=`a``x`2+`k`, where:

- $a$
`a`determines concavity (positive or negative, narrow or wide) - $k$
`k`represents the vertical shift upwards (for $k>0$`k`>0) or downwards (for $k<0$`k`<0)

**4)** $y=\left(x-3\right)^2$`y`=(`x`−3)2 (concave up so $y$`y` has a minimum value at vertex)

- Minimum value of $y$
`y`: $y$`y`is still the result of squaring a value so the smallest possible value is $y=0$`y`=0 - $x$
`x`value corresponding to minimum $y$`y`value: we would need to substitute $x=3$`x`=3 to get $y=0$`y`=0 - Vertex: $\left(3,0\right)$(3,0)

$y=\left(x-3\right)^2$`y`=(`x`−3)2 is the result of a $3$3 unit translation of $y=x^2$`y`=`x`2 to the * right*.

**5)** $y=\left(x+3\right)^2$`y`=(`x`+3)2 (concave up so $y$`y` has a minimum value at vertex)

- Minimum value of $y$
`y`: $y$`y`is still the result of squaring a value so the smallest possible value is $y=0$`y`=0 - $x$
`x`value corresponding to minimum $y$`y`value: here we would need to substitute $x=-3$`x`=−3 to get $y=0$`y`=0 - Vertex: $\left(-3,0\right)$(−3,0)

$y=\left(x+3\right)^2$`y`=(`x`+3)2 is the result of $3$3 unit translation of $y=x^2$`y`=`x`2 to the * left*.

**6)** $y=-\left(x-4\right)^2$`y`=−(`x`−4)2 (concave down so $y$`y` has a maximum value at vertex)

- Maximum value of $y$
`y`: $y$`y`is the result of squaring a value, and then multiplying that square value by $-1$−1, so the largest possible value is $y=0$`y`=0 - $x$
`x`value corresponding to maximum $y$`y`value: here we would need to substitute $x=4$`x`=4 to get $y=0$`y`=0 - Vertex: $\left(4,0\right)$(4,0)

$y=-\left(x-4\right)^2$`y`=−(`x`−4)2 is the result of a $4$4 unit translation of $y=x^2$`y`=`x`2 to the * right*.

Now that we know how to represent vertical and horizontal translations algebraically, we can summarise the key concepts in the form $y=a\left(x-h\right)^2+k$`y`=`a`(`x`−`h`)2+`k`, where:

- $a$
`a`determines concavity - $k$
`k`represents the vertical shift upwards or downwards - $h$
`h`represents the horizontal shift to the right or to the left

In this form, every parabola can be thought of as some sort of transformation of $y=x^2$`y`=`x`2. All we need to consider by looking at the equation is:

- Is there any narrowing, widening or reflecting of the curve $y=x^2$
`y`=`x`2? - Is there any horizontal or vertical translation of the curve $y=x^2$
`y`=`x`2?

A parabola of the form $y=ax^2$`y`=`a``x`2 goes through the point $\left(2,-4\right)$(2,−4).

What is the value of $a$

`a`?What are the coordinates of the vertex?

Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

Plot the graph of the parabola.

Loading Graph...

Consider the parabola $y=8+3\left(x-7\right)^2$`y`=8+3(`x`−7)2.

What is the vertex of the parabola? Give your answer in the form $\left(h,k\right)$(

`h`,`k`).The parabola is reflected about the $y$

`y`-axis. What will its new equation be?What will be the vertex of the new parabola formed after the reflection?

Answer the following.

Write down the equation of the new parabola when $y=x^2$

`y`=`x`2 is translated $6$6 units to the right and $2$2 units down.Graph the parabola formed after the translations.

Loading Graph...

A skydiving instructor wants to use the equation $y=at^2+c$`y`=`a``t`2+`c` to model the height in metres of a skydiver above the ground $t$`t` seconds after jumping out of the plane.

Which of the following would be the correct values of $a$

`a`and $c$`c`to use?Negative $a$

`a`, positive $c$`c`APositive $a$

`a`, negative $c$`c`BNegative $a$

`a`, negative $c$`c`CPositive $a$

`a`, positive $c$`c`DShe does a test run, jumping out of the plane at a height of $2560$2560 metres. Find the value of $c$

`c`.After $160$160 seconds, the instructor lands on the ground. Form an equation to find the value of $a$

`a`.After $23$23 seconds, how far above the ground was she?