Quadratic Functions: Graphs and Properties

Try this

Drag only $c$ . Does the shape of the parabola change? Which point in the picture always follows $c$ ?
Drag only $a$ : from $2$ to $0.5$ , then to $- 1$ . How do the direction of opening and the “width” change? What happens when $a$ is exactly $0$ ?
Drag only $b$ and watch the vertex. What kind of path does it follow? When the axis of symmetry is to the left of the $y$ -axis, how are the signs of $a$ and $b$ related?
Find values of $a$ , $b$ , $c$ that make the discriminant $Δ$ exactly $0$ . How does the graph sit against the $x$ -axis then?
The two diamond points in the picture are at $x = 1$ and $x = - 1$ . Why are their heights exactly $a + b + c$ and $a - b + c$ ?

Predict first

A parabola $y = a x^{2} + b x + c$ opens downward, its axis of symmetry is to the right of the $y$ -axis, and it meets the $y$ -axis above the origin. How many of $a$ , $b$ , $c$ are positive?

0

1

2

3

Choose an answer first; then you can see the result.

Why: completing the square finds the vertex

Every quadratic graph has the same “width” as $y = a x^{2}$ ; it has just been moved. Moved where? Completing the square tells us.

Example: where is the vertex of y = 2x² − 4x − 1?

Step 1 / 5

Start with the simplest one
$y = 2 x^{2}$ : the vertex is at the origin and the axis of symmetry is the $y$ -axis. We want to see how $y = 2 x^{2} - 4 x - 1$ differs from it.

$y = 2 x^{2}$
Take out the leading coefficient
Complete the square inside the bracket
Move the extra bit outside the bracket
Read off the vertex and the axis

Doing the same to a general $y = a x^{2} + b x + c$ gives

y = a (x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a}, axis x = - \frac{b}{2 a} .

No need to cram the formula: if you can complete the square, you can rebuild it any time.

Six signs, six things to look at

Tests often show a graph and ask for the signs of a list of expressions. Each expression has its own “thing to look at” in the picture.

Reading the graph: y = x² + 3x + 1

Step 1 / 6

a: look at the opening
It opens upward, so $a > 0$ . The larger $∣ a ∣$ , the narrower the parabola.

$a > 0$
c: look where it meets the y-axis
b: look which side the axis is on
Δ: count the crossings with the x-axis
a + b + c: look at the point at x = 1
a − b + c: look at the point at x = −1

The same idea reads off even more: $4 a + 2 b + c$ is the value of the function at $x = 2$ , and the sign of $2 a + b$ only needs a comparison between the axis and $x = 1$ . When you see a combination of coefficients, first ask: “the value of the function at which $x$ is this?”

One problem, several solutions

Find the vertex of the parabola $y = x^{2} - 6 x + 5$ .

Complete the square

$y = x^{2} - 6 x + 9 - 9 + 5 = (x - 3)^{2} - 4$ . The vertex is $(3, - 4)$ .

CostAlways works; watch the signs.

Use symmetry

$x^{2} - 6 x + 5 = (x - 1) (x - 5)$ , so the graph meets the $x$ -axis at $x = 1$ and $x = 5$ . The axis is exactly halfway between them: $x = \frac{1 + 5}{2} = 3$ . Substitute: $y = 9 - 18 + 5 = - 4$ . The vertex is $(3, - 4)$ .

CostLeast effort, but the two crossings must be easy to find.

Use the formula

$x = - \frac{b}{2 a} = - \frac{- 6}{2} = 3$ ; substituting gives $y = - 4$ . The vertex is $(3, - 4)$ .

CostFast; get one sign of the formula wrong and everything is wrong.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: The graph of $y = a x^{2} + b x + c$ is a parabola with a highest or lowest point

When $a \neq = 0$ , however $b$ and $c$ change, the graph always has a vertex: the lowest point if it opens upward, the highest point if it opens downward.

Practice

Exercise 1Basic

Find the minimum value of $y = x^{2} - 4 x + 1$ .

Exercise 2Basic

The parabola $y = - 2 (x - 1)^{2} + 8$ meets the $x$ -axis at two points. Give the $x$ -coordinates of these two points.

Exercise 3Advanced

The parabola $y = x^{2} + b x + c$ has vertex $(2, - 1)$ . Write the expression after $y =$ (expanded).

Exercise 4Advanced

A parabola passes through $(- 1, 0)$ , $(3, 0)$ and $(0, 6)$ . Write the expression after $y =$ .

Exercise 5Challenge

As $b$ runs through all real numbers, the vertex of the parabola $y = x^{2} + b x + 1$ moves. All these vertices lie on another parabola. Write the expression after $y =$ for that parabola.

Try this

Why: completing the square finds the vertex

Example: where is the vertex of y = 2x² − 4x − 1?

Six signs, six things to look at

Reading the graph: y = x² + 3x + 1

Complete the square

Use symmetry

Use the formula

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Flipping the sign of the vertex's x-coordinate

Forgetting the coefficient outside the bracket when completing the square