# Quadratic Functions

**4 Graph**

Some quadratic functions are shown in figure 1:

Figure 1: Some Quadratic Functions

Looking at the various graphs we have, we can draw several conclusions.

• There seems to be two kind of graphs. Some open up, the
others open

down.

• All the graphs seem to have a similar shape, that of a U. Besides the way

they open, another important factor in characterizing these graphs seems

to be the position of the graphs with respect to both axes. Some graphs

intersect the x-axis twice, others once, some never. All graphs intersect

the y-axis once.

It turns out that it is fairly easy to predict how the graph will look if we

have its equation. This is what we study next.

**4.1 Orientation**

• If a > 0 the parabola opens up.

• If a < 0 the parabola opens down.

**Example 15 **The graph of y = x^{2} − 4 opens up.

**Example 16** The graph of y + 2x^{2} − 5x = 3 opens down.

4.2 The y-intercept

Remember, to find the y-intercept, we set x = 0 and solve for y. If we do this

in y = ax^{2} +bx+c, we have that y = c. So, the y-intercept of y = ax^{2} +bx+c

is the point (0, c). This tells us that when the equation is in standard form,
we

get the y-intercept with no computations.

**Example 17** The y-intercept of y = x^{2} − 4 is (0,−4)

**Example 18** The y intercept of y +2x^{2} − 5x = 3 is (0, 3)

**4.3 X-intercepts**

Remember, to find the x-intercepts, we set y = 0 and solve for x. To find the

x-intercepts of y = ax^{2} + bx + c, we need to solve ax^{2} + bx + c = 0. This is a

quadratic equation, its solutions are given by the quadratic formula.

**Example 19** Find the x-intercepts of y = x^{2} − x − 6.

To find them, we need to solve x^{2} − x −6 = 0. We already solved this equation

above, and found that its solutions were x = −2 and x = 3. Therefore, the

x-intercepts are (−2, 0) and (3, 0).

**Example 20 **Find the x-intercepts of y = x^{2} + x+ 1.

To find them, we need to solve x^{2} + x +1 = 0. We already solved this equation

above, and found that it had no solutions. Therefore, y = x^{2} + x + 1 has no
xintercepts.

We can even tell more about it. Since the coefficient of x^{2} is positive,

it means that its graph opens up. Hence, the graph is entirely above the x-axis.

**4.4 Vertex**

**Definition 21 (vertex) **The vertex is the highest point on a parabola if the

parabola opens down, it is the lowest point if the parabola opens up. See Figure

2. The vertex is a point, hence finding the vertex means finding its x and y-

Figure 2: Vertex of Quadratic Functions

coordinates.

• The x-coordinate of the vertex is .

• To find the y-coordinate of the vertex, simply plug the x-coordinate in

the equation and compute y. If the quadratic function is y = f (x) =

ax^{2} + bx + c, then the coordinates of the vertex are
.

• If the quadratic function is in vertex form, then the coordinates of the

vertex are (h, k).

**Example 22 **Find the vertex of y = x^{2} + x + 1.

Here, a = b = c = 1. Hence, the x-coordinate of the vertex
is

The y-coordinate of the vertex is then

The coordinates of the vertex are:
. In this case, knowing the vertex

also provides information about the x-intercepts. Since this parabola opens up,

and its vertex (lowest point) is above the x-axis, the entire parabola has to be

above the x-axis. Hence, it cannot have x-intercepts.

**Example 23** Find the vertex of y = x^{2} − 8x+ 16.

Here, a = 1, b = −8, c = 16. The x-coordinate of the vertex is
.

The y-coordinate of the vertex is y = 4^{2} − (8) (4) + 16 = 0. So, the vertex is

(4, 0). Note that since the y-coordinate of the vertex is 0, the vertex is on
the

x-axis, hence it is also an x-intercept.

**Example 24 **Find the vertex of y = 2(x − 3)^{2} + 5

This function is in vertex form, with h = 3 and k = 5. The vertex is (3, 5).

**
Remark 25** If a parabola opens up (a > 0) and its vertex is above the x-axis

(its y-coordinate > 0) then the parabola has no x-intercepts.

**Remark 26**If a parabola opens down (a < 0) and its vertex is below the x-axis

(its y-coordinate < 0) then the parabola has no x-intercepts.

**We can combine the previous two remarks into one which states:**

Remark 27

Remark 27

If a and the y-coordinate of the vertex have the same sign, the corresponding

parabola has no x-intercepts.

**Remark 28**Finally, if the vertex is on the x-axis (its y-coordinate is 0) then

there is only one x-intercept, the same as the vertex.

The above remarks tell us that we should always find the vertex before the

x-intercepts. In some cases, knowing the vertex allows us to draw conclusions

about the x-intercept.

**4.5 Technology Note**

I have developed several applets to experiment with quadratic functions:

1. To experiment with the role a, b, and c play in the graph of y = ax^{2}+bx+c.

2. To experiment with the role a, h and k play in the
graph of y = a (x − h)^{2}+

k.

3. To experiment with a graphing applet which can graph
any function, similar

to the TI 83.

**5 Maximum, Minimum and Range**

Consider the quadratic function y = f (x) = ax^{2} + bx + c. Remember that the

y-coordinate of a point gives us the vertical position of that point. When a

parabola opens up, its vertex is the lowest point, hence its y-coordinate
corresponds

to the smallest value y (i.e. the quadratic function) can have. Similarly,

if the parabola opens down, the y-coordinate of its vertex corresponds to the

largest value y (i.e. the quadratic function) can have. We summarize this by:

• If a > 0, the quadratic function f (x) = ax^{2} + bx + c has no maximum.

Its minimum is
(the y-coordinate of the vertex) . Its range is

• If a < 0, the quadratic function f (x) = ax^{2} + bx + c has no minimum.

Its maximum is (the y-coordinate of the vertex) . Its range is

So, questions related to finding the maximum or minimum of a quadratic

function are solved by finding the vertex of the function.

**Example 29** Find the maximum, the minimum and the range of f (x) = −x^{2}+

4x + 2.

Here, a = −1, b = 4, c = 2. The x-coordinate of the vertex is
.

The y-coordinate is −2^{2} +4(2)+2 = 6. Hence the vertex is (2, 6). Since the its

graph opens down, f (x) has no minimum. Its maximum is f (x) = 6. Its range

is all reals less than or equal to 6.