Quadratic Range and Restricted Domains | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

When the curve opens downward

Worked example

Find the maximum point of $y=16x-7-4x^2$

Complete the square. Taking out $-4$ gives the following form.
$y=9-4(x-2)^2$
The squared part is never negative, so subtracting it cannot make the value larger than $9$ . The largest $y$ can be is $9$ , and that happens when the bracket is zero.
$x-2=0 \;\Rightarrow\; x=2$

Answer

\text{Maximum point } (2,\,9)

The range

The is the set of $y$ -values the curve actually reaches.
For an upward parabola the vertex is the lowest point, so the range starts at $k$ and goes up: $f(x)\ge k$ .
For a downward parabola the vertex is the highest point, so the range is $f(x)\le k$ .

Worked example

The function $f$ is defined by $f(x)=2x^2+4x+10$ for all real $x$ . State the range. (9709/13/O/N/25 Q10(a))

Complete the square first.
$f(x)=2(x+1)^2+8$
Here $a=2>0$ , so the vertex is a minimum and the least value is $8$ .

Answer

f(x)\ge 8

Common mistake

Write the range using

f(x)

y

, never

x

. The range describes the output of the function;

x

is the input. In this question, the required statement is

f(x)\ge8

, not

x\ge8

Your turn9709/13/M/J/25 Q11(a)

The function $f$ is defined by $f(x)=x^2+4ax+a$ for all real $x$ , where $a$ is a constant. Given that the range of $f$ is $f(x)\ge-33$ , find the possible values of $a$ .

Show worked answer

Complete the square while carrying the letter $a$ . The least value is the constant at the end, so set it equal to $-33$ .

f(x)=(x+2a)^2-4a^2+a

-4a^2+a=-33

4a^2-a-33=0

(4a+11)(a-3)=0

Answer

a=-\tfrac{11}{4} \quad\text{or}\quad a=3

When the domain removes the vertex

One paper defined $f(x)=-2x^2-8x-13$ only for $x<-3$ (9709/13/O/N/22 Q2). Completing the square gives $f(x)=-2(x+2)^2-5$ , so the vertex is at $(-2,-5)$ .

But $x=-2$ is not allowed, because the domain only includes $x<-3$ . The curve never reaches its vertex.
On $x<-3$ the curve is still rising towards the vertex, so the values get closest to $f(-3)=-7$ without ever reaching it.
So the range is $f(x)<-7$ . Whenever a domain is given, check whether the vertex is inside it before you write the range.

When the curve opens downward

Worked example

Find the maximum point of $y=16x-7-4x^2$

Complete the square. Taking out $-4$ gives the following form.
$y=9-4(x-2)^2$
The squared part is never negative, so subtracting it cannot make the value larger than $9$ . The largest $y$ can be is $9$ , and that happens when the bracket is zero.
$x-2=0 \;\Rightarrow\; x=2$

Answer

\text{Maximum point } (2,\,9)

The range

The is the set of $y$ -values the curve actually reaches.
For an upward parabola the vertex is the lowest point, so the range starts at $k$ and goes up: $f(x)\ge k$ .
For a downward parabola the vertex is the highest point, so the range is $f(x)\le k$ .

Worked example

The function $f$ is defined by $f(x)=2x^2+4x+10$ for all real $x$ . State the range. (9709/13/O/N/25 Q10(a))

Complete the square first.
$f(x)=2(x+1)^2+8$
Here $a=2>0$ , so the vertex is a minimum and the least value is $8$ .

Answer

f(x)\ge 8

Common mistake

Write the range using

f(x)

y

, never

x

. The range describes the output of the function;

x

is the input. In this question, the required statement is

f(x)\ge8

, not

x\ge8

Your turn9709/13/M/J/25 Q11(a)

The function $f$ is defined by $f(x)=x^2+4ax+a$ for all real $x$ , where $a$ is a constant. Given that the range of $f$ is $f(x)\ge-33$ , find the possible values of $a$ .

Show worked answer

Complete the square while carrying the letter $a$ . The least value is the constant at the end, so set it equal to $-33$ .

f(x)=(x+2a)^2-4a^2+a

-4a^2+a=-33

4a^2-a-33=0

(4a+11)(a-3)=0

Answer

a=-\tfrac{11}{4} \quad\text{or}\quad a=3

When the domain removes the vertex

One paper defined $f(x)=-2x^2-8x-13$ only for $x<-3$ (9709/13/O/N/22 Q2). Completing the square gives $f(x)=-2(x+2)^2-5$ , so the vertex is at $(-2,-5)$ .

But $x=-2$ is not allowed, because the domain only includes $x<-3$ . The curve never reaches its vertex.
On $x<-3$ the curve is still rising towards the vertex, so the values get closest to $f(-3)=-7$ without ever reaching it.
So the range is $f(x)<-7$ . Whenever a domain is given, check whether the vertex is inside it before you write the range.