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Stretching, shrinking and the absolute value

Stretching, shrinking and the absolute value

Stretching and shrinking

Multiplying a function by a number multiplies every value by it, and the graph is stretched or shrunk along the y\displaystyle y-axis:

y=a⋅f(x)\displaystyle y=a\cdot f(x)
  • ∣a∣>1\displaystyle |a|>1 – a stretch: the graph moves away from the x\displaystyle x-axis.
  • ∣a∣<1\displaystyle |a|<1 – a shrink: the graph moves closer to the x\displaystyle x-axis.
  • a<0\displaystyle a<0 – a reflection in the x\displaystyle x-axis as well.

On the board: f(x)=x2\displaystyle {f(x)}={x^2}, g(x)=2x2\displaystyle {g(x)}={2x^2} and h(x)=12x2\displaystyle {h(x)}={\frac{1}{2}x^2} (the dashed ones).

f(x)=x2\displaystyle f(x)=x^{2}
g(x)=2x2\displaystyle g(x)=2x^{2}
h(x)=12x2\displaystyle h(x)=\frac{1}{2}x^{2}
(1,1)\displaystyle (1,1)
(1,2)\displaystyle (1,2)
  • The point (1,1)\displaystyle (1,1) of f\displaystyle f moves to (1,2)\displaystyle (1,2) on g\displaystyle g: the x\displaystyle x stays, the y\displaystyle y is doubled.
  • Every point (x0,y0)\displaystyle (x_0,y_0) of f\displaystyle f moves to (x0, a⋅y0)\displaystyle (x_0,\ a\cdot y_0).
  • The points where the graph meets the x\displaystyle x-axis do not move: 0\displaystyle 0 times any number is 0\displaystyle 0.
The absolute value

The absolute value of a number is its distance from 0 on the number line: ∣5∣=5\displaystyle {|5|}={5}, ∣−5∣=5\displaystyle {|-5|}={5}, ∣0∣=0\displaystyle {|0|}={0}.

∣a∣={aa≥0−aa<0\displaystyle |a|=\begin{cases}a & a\geq0\\-a & a<0\end{cases}

On the board: y=∣x∣\displaystyle {y}={|x|}.

y=∣x∣\displaystyle y=|x|
  • To the right of 0 the graph is y=x\displaystyle {y}={x}; to the left of 0 it is y=−x\displaystyle {y}={-x}.
  • An absolute value is never negative, so the whole graph is above the x\displaystyle x-axis or on it.
The graph of y=∣f(x)∣\displaystyle {y}={|f(x)|}

In the graph of y=∣f(x)∣\displaystyle {y}={|f(x)|} everything above the x\displaystyle x-axis stays, and everything below it is reflected above it.

On the board: f(x)=x2−4\displaystyle {f(x)}={x^2-4} (the dashed one) and g(x)=∣x2−4∣\displaystyle {g(x)}={|x^2-4|}.

f(x)=x2−4\displaystyle f(x)=x^{2}-4
g(x)=∣x2−4∣\displaystyle g(x)=|x^{2}-4|
(0,4)\displaystyle (0,4)
  • Between x=−2\displaystyle {x}={-2} and x=2\displaystyle {x}={2} the values of f\displaystyle f are negative, and there the graph of g\displaystyle g is their reflection: g(x)=−f(x)\displaystyle {g(x)}={-f(x)}.
  • The minimum point of f\displaystyle f, (0,−4)\displaystyle (0,-4), moves to (0,4)\displaystyle (0,4): a maximum point of g\displaystyle g.
  • At the points (−2,0)\displaystyle (-2,0) and (2,0)\displaystyle (2,0) the value of g\displaystyle g is 0, the least it has: these are the minimum points of g\displaystyle g.
Several transformations together

g(x)=2(x−1)2−3\displaystyle {g(x)}={2(x-1)^2-3}.
The example below gets the graph of g\displaystyle g from the graph of f(x)=x2\displaystyle {f(x)}={x^2} in three steps.

  • A stretch by 2: y=2x2\displaystyle {y}={2x^2}.
  • A shift of one unit to the right: y=2(x−1)2\displaystyle {y}={2(x-1)^2}.
  • A shift of 3 units down: y=2(x−1)2−3\displaystyle {y}={2(x-1)^2-3}.
f(x)=x2\displaystyle f(x)=x^{2}
g(x)=2(x−1)2−3\displaystyle g(x)=2(x-1)^{2}-3
V=(1,−3)\displaystyle V=(1,-3)
  • The vertex (0,0)\displaystyle (0,0) does not move in the stretch, and the two shifts take it to (1,−3)\displaystyle (1,-3).
  • The order matters: shifting 3 down first and stretching afterwards gives 2(x2−3)=2x2−6\displaystyle {2(x^2-3)}={2x^2-6} – another function.
Exercise 1

On the board: f(x)=x2\displaystyle {f(x)}={x^2}, and two graphs whose formulas are hidden.
Each of them is a⋅f(x)\displaystyle a\cdot f(x), for another a\displaystyle a.
Write the formula of each graph in a row of its own.

f(x)=x2\displaystyle f(x)=x^{2}
g(x)= ?\displaystyle g(x)=\,?
h(x)= ?\displaystyle h(x)=\,?

Exercise 2

f(x)=x2−2x−3\displaystyle {f(x)}={x^2-2x-3}.
g(x)=∣x2−2x−3∣\displaystyle {g(x)}={|x^2-2x-3|}.
The minimum point of f\displaystyle f is (1,−4)\displaystyle (1,-4).

a

Mark the points where the graph of g\displaystyle g meets the x\displaystyle x-axis.

f(x)=x2−2x−3\displaystyle f(x)=x^{2}-2x-3
g(x)=∣x2−2x−3∣\displaystyle g(x)=|x^{2}-2x-3|

b

What are the minimum points of g\displaystyle g?
What is the maximum point of g\displaystyle g between the points where it meets the axis?

Exercise 3

f(x)=x\displaystyle {f(x)}={\sqrt{x}}.
The graph of g\displaystyle g is the graph of f\displaystyle f, stretched by 2 and shifted one unit to the right.

a

Write the formula of g\displaystyle g on the board.

f(x)=x\displaystyle f(x)=\sqrt{x}
g(x)= ?\displaystyle g(x)=\,?

b

What is the domain of g\displaystyle g?
At which point does its graph start?

Exercise 4

g(x)=−(x+1)2+4\displaystyle {g(x)}={-(x+1)^2+4}.

a

How do you get the graph of g\displaystyle g from the graph of y=x2\displaystyle {y}={x^2}?
Write the steps in order.

b

Mark the extremum point of g\displaystyle g and the points where its graph meets the x\displaystyle x-axis.

g(x)=−(x+1)2+4\displaystyle g(x)=-(x+1)^{2}+4

Exercise 5

f(∣x∣)\displaystyle f(|x|) means: put ∣x∣\displaystyle |x| in f\displaystyle f in place of x\displaystyle x. For example, if f(x)=x−1\displaystyle {f(x)}={x-1}, then f(∣x∣)=∣x∣−1\displaystyle {f(|x|)}={|x|-1}.
Is the claim true for every function f\displaystyle f?
If so – explain why. If not – give a counterexample.

a

The function ∣f(x)∣\displaystyle |f(x)| is an even function.

b

The function f(∣x∣)\displaystyle f(|x|) is an even function.

c

The graph of y=∣f(x)∣\displaystyle {y}={|f(x)|} passes through the origin.

d

No value of the function ∣f(x)∣\displaystyle |f(x)| is negative.