Quadratic Equations

New Zealand

Level 7 - NCEA Level 2

Lesson

We've already learnt how to find certain features of linear equations including the gradient, the $x$`x`-intercept, and the $y$`y`-intercept. Now we are going to explore the features parabolas.

The intercepts are where the parabola crosses the horizontal ($x$`x`) and vertical ($y$`y`) axes.

More specifically:

- The $x$
`x`-intercepts are where the parabola crosses the $x$`x`-axis. This occurs when $y=0$`y`=0. - The $y$
`y`-intercept is where the parabola crosses the $y$`y`-axis. This occurs when $x=0$`x`=0.

You can see them in the picture below. Note, however, that there won't always be two $x$`x`-intercepts. Sometimes there may only be one or even none.

Now let's look at the process of finding the $x$`x`- and $y$`y`-intercepts of a quadratic equation. We'll use the equation $y=x^2-2x-3$`y`=`x`2−2`x`−3.

To find the $y$`y`-intercept, we substitute $x=0$`x`=0 into the equation. Let's do that now:

$y$y |
$=$= | $x^2-2x-3$x2−2x−3 |

$=$= | $0^2-2\times0-3$02−2×0−3 | |

$=$= | $-3$−3 |

So the $y$`y`-intercept is $\left(0,-3\right)$(0,−3). We can also write that the $y$`y`-value of the $y$`y`-intercept is $-3$−3. Notice that this is where the function crosses the $y$`y`-axis in the image above.

We can also see from this image that the $x$`x`-intercepts are at $x=-1$`x`=−1 and $x=3$`x`=3. To check algebraically that these are the intercepts, we can substitute either value into the equation, and we should expect to end up with $y=0$`y`=0. Let's try with $x=-1$`x`=−1:

$y$y |
$=$= | $x^2-2x-3$x2−2x−3 |

$=$= | $\left(-1\right)^2-2\times\left(-1\right)-3$(−1)2−2×(−1)−3 | |

$=$= | $1+2-3$1+2−3 | |

$=$= | $0$0 |

Success! How about the other intercept, $x=3$`x`=3?

$y$y |
$=$= | $x^2-2x-3$x2−2x−3 |

$=$= | $3^2-2\times3-3$32−2×3−3 | |

$=$= | $9-6-3$9−6−3 | |

$=$= | $0$0 |

So we can see that there are two points of intersection on the $x$`x`-axis that occur at $\left(-1,0\right)$(−1,0) and $\left(3,0\right)$(3,0). Later we will look at other methods for finding the $x$`x`-intercepts of functions that involve factorising.

Maximum or minimum values are also known as the turning points, and they are found at the vertex of the parabola.

Parabolas that are concave up have a minimum value. This means the $y$`y`-value will never go under a certain value.

Parabolas that are concave down have a maximum value. This means the $y$`y`-value will never go *over* a certain value.

Maximum and minimum values occur on a parabola's axis of symmetry. This is the line that evenly divides a parabola into two sides down the middle.

It's easy to see a maximum or minimum value when we look at graphs. However, we can also find these minimum/maximum values algebraically. To do this, we can first find the axis of symmetry:

$x=\frac{-b}{2a}$`x`=−`b`2`a`

Then we can substitute this value of $x$`x` into the quadratic to find the corresponding value of $y$`y`, which will be the maximum or minimum value of the quadratic equation. Remember, the general form is $y=ax^2+bx+c$`y`=`a``x`2+`b``x`+`c`, so we just need to substitute in the relevant values.

For example, let's find the equation of the axis of symmetry for $y=2x^2-4x+7$`y`=2`x`2−4`x`+7. In this example, $a=2$`a`=2 and $b=-4$`b`=−4, so let's substitute them into the formula:

$x$x |
$=$= | $\frac{-\left(-4\right)}{2\times2}$−(−4)2×2 |

$=$= | $\frac{4}{4}$44 | |

$=$= | $1$1 |

So the axis of symmetry for this quadratic has the equation $x=1$`x`=1.

- Determine whether the parabola is concave up or down so we know whether we're finding a minimum or maximum value.
- The turning point will be on the axis of symmetry, so we can find the $x$
`x`-value of this point using the formula $x=\frac{-b}{2a}$`x`=−`b`2`a`. - Substitute the value from step 2 into the original equation to find the $y$
`y`-value.

We have learned about positive and negative gradients when we looked at straight lines. Generally, straight lines had only one or the other.

However, a parabola has a positive gradient in some places and negative gradient in others. The parabola's maximum or minimum value, at which the gradient is $0$0, is the division between the positive and negative gradient regions. Hence why it is also called the turning point.

Look at the picture below. See how one side of the parabola has a positive gradient, then there is a turning point with a zero gradient, then the other side of the parabola has a negative gradient?

We can see in this particular graph that the gradient is positive when $x<1$`x`<1 and we can say that the parabola is increasing for these values of $x$`x`. Similarly, the gradient is negative when $x>1$`x`>1, and for these values of $x$`x` the parabola is decreasing.

Consider the following graph.

Find the coordinates of the $x$

`x`-intercept(s). Give each intercept in the form $\left(a,b\right)$(`a`,`b`).If there is more than one $x$

`x`-intercept, state all of them on the same line, separated by commas.State the zero(s) of the function.

If there is more than one zero, state all of them on the same line separated by commas.

Examine the given graph and answer the following questions.

What are the $x$

`x`values of the $x$`x`-intercepts of the graph? Write both answers on the same line separated by a comma.What is the $y$

`y`value of the $y$`y`-intercept of the graph?What is the minimum value of the graph?

Examine the attached graph and answer the following questions.

What is the $x$

`x`-value of the $x$`x`-intercept of the graph?What is the $y$

`y`value of the $y$`y`-intercept of the graph?What is the absolute maximum of the graph?

Determine the interval of $x$

`x`in which the graph is increasing.Give your answer as an inequality.