## Summary and examples

The general equation of a quadratic function is:

$y=ax^2+bx+c$

$a$, $b$ and $c$ are called parameters. $a\neq0$, because otherwise it is no quadratic function.

The graph of a quadratic function is a parabola. There are two types, one that opens up and one that opens down, depending on the parameter $a$.

$a>0$: opens up

$a<0$: opens down

A parabola intersects the $Y$-axis in the point $(0,c)$. You can easily see this, because for any point on the $Y$-axis we have $x=0$. When $x=0$ is substituted in the equation we get $y=c$.

When a parabola intersects the $X$-axis we can calculate the intersection points in various ways. The $abc$-formula, explained in Quadratic equations (abc-formula), always gives a solution. Another way is based on factorizing the equation, see the topic Quadratic equations (factorizing). This is faster, though not always possible.

The intersection points can be calculated by applying the $abc$-formula:

$x_{1,2}=\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this formula the discrriminant plays an important role:

$D=b^2-4ac$

We have:

• $D>0$: the parabola has two different intersection points with the $X$-axis;
• $D=0$: the parabola has two coinciding intersection points with the $X$-axis (usually it is said that the parabola has just one intersection point); and
• $D<0$: the parabola has no intersection points with the $X$-axis.

A parabola is a symmetric graph and has a symmetry axis, namely the line:

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

Then it is clear that the top of the parabola is on this symmetry axis:

$x_{top}=-\displaystyle\frac{b}{2a}$

This can also be deduced from the $abc$-formula. This optimum can also be found by differentiating the function and make the derivative equal to $0$:

$y'=2ax+b=0$
$x=-\displaystyle\frac{b}{2a}$

#### Example 1

Given the function:

$y=3x^2+2x-1$

1. Determine the intersection point of the graph with the $Y$-axis;
2. Does this graph have an intersection point with the $X$-axis, and if so, how many and determine their coordinates.
3. Determine the symmetry axis.

1. Because $c$ in the formula is equal to $-1$, the intersection point with the $Y$-axis is equal to $(0,-1)$.
2. We have to compute the discriminant:

$D=2^2-4.3.-1=16$

$D$ is greater than $0$ en thus the graph has two intersection points with the $X$-axis. The $x$-coordinates of these intersection points are:

$\displaystyle\frac{-2\pm\sqrt{16}}{2.3}=\displaystyle\frac{-2\pm4}{6}$

so:

$x_1=-1$
$x_2=\displaystyle\frac{1}{3}$

and the coordinates are:

$(-1,0)$  and $(\displaystyle\frac{1}{3},0)$

3. The symmetry axis can be calculated in two way. First by using the formula above, but also by taking the mean value of the two intersection points, namely:

$x=\displaystyle\frac{-1+\displaystyle\frac{1}{3}}{2}=-\displaystyle\frac{1}{3}$

#### Example 2

Given the function:

$y=x^2-6x+8$

1. Determine the intersection point of the graph with the $Y$-axis;
2. Does this graph have an intersection point with the $X$-axis, and if so, how many and determine their coordinates.
3. Determine the symmetry axis.

1. Because $c$ in the formula is equal to $8$, the intersection point with the $Y$-axis is equal to $(0,8)$.
2. We can compute the discriminant to find out whether there are one or more intersection points, but another way is faster. The equation can be factorized:

$y=x^2-6x+8=(x-4)(x-2)$

and thus we may conclude that the graph has two intersection points with the $X$-axis: $(2,0)$ en $(4,0)$.

3. The symmetry axis lies just in the middle of these two points, so the symmetriy-axis is $x=3$.

#### Example 3

For which value of $p$ does the parabola:

$y=x^2-3x+p$

have two coinciding intersection points with the $X$-axis.

These intersection points coincide if the discriminant $D=0$, so:

$D=9-4p=0$
$p=\displaystyle\frac{9}{4}$

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