Quadratic with zeros 3 and -1 through (1, -8) Step by Step

Step by step

The question

Question: write a quadratic with zeros 3 and −1 through (1, −8).

zeros : 3, −1

through : (1, −8)

A zero of f is an x that makes f(x) equal 0.

zero: an x where f(x) = 0

1 Each zero gives a bracket

If c is a zero, then x minus c is a factor.

Put c in, and that bracket is 0, so all of f(x) is 0.

zero c → bracket (x − c)

The zero 3 gives x − 3.

x = 3 → x − 3

Next, −1 gives x − (−1).

Minus a minus is plus: x + 1.

x = −1 → x − (−1) = x + 1

Careful!

The sign flips.

The zero 3 gives x − 3, because that is 0 when x is 3.

Careful: not zero 3 → (x + 3) ✗

2 Write f(x)

The zeros fix the brackets, but f can be stretched: a times the brackets.

The point tells us a.

f(x) = a(x − 3)(x + 1)

3 Find a

Put x = 1 and f(x) = −8 in.

−8 = a(1 − 3)(1 + 1)

Work out each bracket: −2 and 2.

−8 = a(−2)(2)

−2 times 2 is −4.

So −8 = −4a.

−8 = −4a

To remove times −4, we divide both sides by −4.

−8−4 = −4a−4

−4 divided by −4 is 1.

So just a is left.

So a = 2.

a = 2

We put a in front of the brackets.

f(x) = 2(x − 3)(x + 1)

4 Multiply out

Each part of the first bracket times each part of the second.

each part × each part

(x − 3) times (x + 1).

Each part times each part.

= 2[x² + x − 3x − 3]

Like terms together: x² − 2x − 3.

= 2(x² − 2x − 3)

2 times each part.

= 2x² − 4x − 6

So f(x) = 2x² − 4x − 6.

f(x) = 2x² − 4x − 6

5 Check

Let's check: put each zero in.

First x = 3.

We get 0.

f(3) = 2(3)² − 4(3) − 6 = 18 − 12 − 6 = 0

And x = −1.

We get 0.

f(−1) = 2(−1)² − 4(−1) − 6 = 2 + 4 − 6 = 0

And the point: put x = 1 in.

We get −8, as it should be.

f(1) = 2(1)² − 4(1) − 6 = 2 − 4 − 6 = −8

Let's look back.

First: each zero gives a bracket.

Then: write f(x).

Then: find a.

Then: multiply out.

Last: check.

Answer