A. Characteristics of Functions · Quadratic Functions

Using the Discriminant: Unknown Coefficients and "Can It Reach?"

A2.1 A2.3About 65 minutes

Learning goals

I can find values of a parameter that give a quadratic function two, one or no zeros.
I can solve simple equations and inequalities that come from the discriminant.
I can decide whether a quadratic function ever reaches a given value.
I can connect D = 0 to the maximum or minimum value.

Words to know

parameter
A letter such as k that stands for a fixed but unknown number in an equation, as in f(x) = 2x² − 4x + k.

condition
An equation or inequality that must hold, such as D = 0 for exactly one zero.

Working backwards

In the last lesson you used the discriminant to count zeros. Now turn the question around. A designer wants a parabola that just touches the ground. A coach wants to know if a ball can ever reach a certain height. A business asks whether its revenue can possibly reach a target. In each case there is an unknown in the equation, and the discriminant tells you which values work.

The three conditions

For f(x) = ax² + bx + c with D = b² − 4ac:

two zeros ⇔ D > 0
one zero ⇔ D = 0
no zeros ⇔ D < 0

If a, b or c contains an unknown such as k, write D in terms of k, then solve the equation D = 0 or the inequality D > 0 or D < 0. (⇔ means "exactly when".)

Worked example

Example 1: Exactly one zero

Find all values of k for which f(x) = x² + kx + 9 has exactly one zero.

a = 1, b = k, c = 9, so D = k² − 4(1)(9) = k² − 36.
One zero: k² − 36 = 0, so k² = 36 and k = ±6.
Check: x² + 6x + 9 = (x + 3)² and x² − 6x + 9 = (x − 3)² are perfect squares ✓.

k = 6 or k = −6.

Worked example

Example 2: All three cases

For f(x) = 2x² − 4x + k, find the values of k that give two zeros, one zero and no zeros.

D = (−4)² − 4(2)(k) = 16 − 8k.
Two zeros: 16 − 8k > 0, so 16 > 8k and k < 2.
One zero: 16 − 8k = 0, so k = 2.
No zeros: 16 − 8k < 0, so k > 2.
This makes sense: changing k slides the parabola up and down (a vertical translation).

Two zeros for k < 2, one for k = 2, none for k > 2.

Members of f(x) = 2x² − 4x + k. Increasing k slides the parabola up. At k = 2 the vertex (1, 0) sits on the x-axis.

The values of k for f(x) = 2x² − 4x + k on a number line. The single value k = 2 separates two zeros from no zeros.

Worked example

Example 3: Unknown in the leading coefficient

For what values of k does kx² + 6x + 3 have no real zeros?

D = 36 − 4(k)(3) = 36 − 12k.
No zeros: 36 − 12k < 0, so 36 < 12k and k > 3.
(For k > 3 the function is quadratic, since k ≠ 0.)

k > 3.

Worked example

Example 4: Can the ball get that high?

A volleyball is served so that its height in metres after t seconds is h(t) = −4.9t² + 20t + 1.5. Can it reach 25 m?

Ask whether h(t) = 25 has a solution: −4.9t² + 20t + 1.5 = 25, so −4.9t² + 20t − 23.5 = 0.
D = 20² − 4(−4.9)(−23.5) = 400 − 460.6 = −60.6.
D < 0, so the equation has no real solutions: the ball never reaches 25 m.
Confirm: the maximum is at t = −20/(2 × (−4.9)) ≈ 2.04 s, and h ≈ 1.5 + 400/19.6 ≈ 21.9 m, less than 25 m ✓.

No. Its maximum height is only about 21.9 m.

Setting up a "can it reach" question

To ask "does f(x) ever equal m?", rearrange f(x) = m into the form ax² + bx + c = 0 and check its discriminant. D > 0: twice. D = 0: exactly once, at the vertex, so m is the maximum or minimum value. D < 0: never.

Hands-on activity

Discriminant dice

You need: two or three dice (or a deck of cards with face cards removed, so the values are 1 to 10), paper and a pencil. Graphing technology (such as a free online graphing calculator) is optional for checking.

Roll to get a and c. Write f(x) = ax² + kx + c with k unknown.
Find every value of k that gives exactly one zero (solve k² − 4ac = 0).
Pick one of your k-values and factor f(x) as a perfect square to check.
Play with a partner: the first to find both k-values correctly scores a point. Play to five.

What to look for: k = ±2√(ac). For example, a = 4 and c = 9 gives k² = 144, so k = ±12, and 4x² + 12x + 9 = (2x + 3)². If ac is not a perfect square, the answers are radicals, such as a = 2, c = 3: k = ±2√6.

Just for fun

Teacher: "Will this ball ever reach 25 m?" Student: "Let me ask the discriminant." Discriminant: "It's negative. The ball isn't going anywhere near that height, and honestly neither am I."

Your turn

Find k so that x² + kx + 16 has exactly one zero.

For what values of c does x² − 6x + c have two zeros?

For what values of k does 3x² + 2x + k have no zeros?

Find k so that kx² − 4x + 1 has exactly one zero.

For what values of k does f(x) = −x² + 4x + k have two zeros?

h(t) = −5t² + 20t + 1 gives a ball's height in metres. Does it reach 22 m?

Does the same ball reach 21 m? How many times?

More practice: warm-up

Write the discriminant of x² + kx + 4 in terms of k.

For what value of c does x² + 10x + c have one zero?

A discriminant is D = k − 5. For what values of k are there two zeros?

Is x² + 2x + 5 ever equal to zero?

Which inequality must D satisfy for no real zeros?

More practice: on your own

Find k so that 4x² + kx + 1 has one zero.

For what values of k does x² + kx + 4 have two zeros?

2x² + 3x + k = 0 has two real roots. What is the greatest integer value of k?

A ticket price p (in dollars) gives revenue R(p) = −50p² + 1000p dollars. Can the revenue reach $6000?

A snowboard jump's path is h(x) = −0.1x² + 1.2x + 0.5 metres, where x is the horizontal distance in metres. Does the rider reach 4.5 m?

Challenge

For what values of k does (k − 1)x² + 4x + 2 = 0 have exactly one real root?

Show that x² + kx − 1 = 0 has two real roots for every value of k.

Find all values of m for which f(x) = x² + mx + m + 3 has exactly one zero, and check each.

Check your understanding

Quiz

For what values of k does x² + kx + 25 have exactly one zero?

For what values of c does 3x² − 6x + c have no zeros?

h(t) = −5t² + 30t + 1 is a ball's height in metres. Does it reach 50 m?

f(x) = x² + 4x + k has two zeros when:

The discriminant of x² + kx − 3 is k² + 12. What can you conclude?

kx² + 2x + 1 has exactly one zero (with k ≠ 0) when:

For parents and teachers

Big idea: Still expectation A2.1 (number of zeros), now used in reverse to find unknown coefficients and to answer "can it ever reach…?" questions (A2.3).

How to help: Ask your teen to say in words what D > 0, D = 0 and D < 0 mean before they calculate.

Common difficulties:

Solving k² > 16 as k > 4 only. The answer is k < −4 or k > 4.
Flipping (or failing to flip) the inequality sign when dividing by a negative number.
Forgetting to rearrange to "= 0" before finding a, b and c in a "can it reach" question.
Forgetting that a coefficient of x² equal to 0 makes the equation linear.