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Interpret and determine a transformed power function

Start a set of 8
Step 1 of 5

Which information should be read first from the transformed power rule?

\[y=-2(x-3)^4+5\]
Year 11, VCE Units 1 & 2, Chapter 5, 4 difficulty levels, No calculatorBuild a worksheet

Transformed power functions

The form y=a(x-h)^n+k shows the centre, exponent and scale of a transformed power function. Read those features first, then set y to zero when finding horizontal-axis intercepts

\[y=a\left(x-h\right)^n+k\]
  1. Read the centre In y=a(x-h)^n+k, the centre has coordinates (h,k) \(\displaystyle y=2(x-3)^4+1\Rightarrow(h,k)=(3,1)\)
  2. Use exponent parity For a nonzero scale a, an even integer exponent n ≥ 2 gives line symmetry and a turning point, while an odd integer exponent n ≥ 3 gives point symmetry and a stationary point of inflection \(\displaystyle n=4\Rightarrow\text{line symmetry about }x=h,\qquad n=5\Rightarrow\text{point symmetry about }(h,k)\)
  3. Use the scale sign The sign of a determines the orientation after the exponent parity is considered \(\displaystyle y=-3(x-1)^4+2\Rightarrow\text{both ends fall}\)
  4. Solve the intercept equation Set y equal to zero, isolate the power and use all real branches \(\displaystyle 0=2(x-1)^4-32\Rightarrow(x-1)^4=16\Rightarrow x=-1,\ 3\)

Worked example

Determine the transformed power rule with exponent 5 and centre (2,3) that passes through (3,5)

\[n=5,\qquad(h,k)=(2,3),\qquad(3,5)\text{ lies on the graph}\]
  1. \(\displaystyle y=a(x-2)^5+3\)Use the transformed power form with the given exponent and centre
  2. \(\displaystyle 5=a(3-2)^5+3\)Substitute the given point into the rule
  3. \(\displaystyle 2=a(1)^5\Rightarrow a=2\)Subtract 3 and divide by the translated power to find the scale
  4. \(\displaystyle y=2(x-2)^5+3\)The recovered rule has odd-power point symmetry about (2,3)

Common mistakes

Reversing the horizontal shift
✕\(\displaystyle y=2(x-3)^4-32\Rightarrow h=-3\)✓\(\displaystyle y=2(x-3)^4-32\Rightarrow h=3\)
The bracket is x-h, so x-3 gives h=3.
Ignoring exponent parity
✕\(\displaystyle n=4\Rightarrow\text{point symmetry and a stationary point of inflection}\)✓\(\displaystyle n=4\Rightarrow\text{line symmetry and a turning point}\)
An even exponent produces line symmetry and a turning point at the centre
Reversing the scale orientation
✕\(\displaystyle y=-2(x-1)^4+2\Rightarrow\text{both ends rise}\)✓\(\displaystyle y=-2(x-1)^4+2\Rightarrow\text{both ends fall}\)
A negative scale reflects an even-power graph so both ends fall away from the centre

Difficulty levels

introIdentify parity, centre and orientation from a canonical transformed power rule
standardCalculate horizontal intercepts and choose a graph for a translated or dilated power
challengeRecover a transformed power rule from its centre and one exact point or compare family members
examUse an inner-scaled rule to complete feature, graph and exact verification steps

Questions about this topic

What are h and k in y=a(x-h)^n+k?+

They give the centre coordinates (h,k). The horizontal coordinate is read with the opposite sign from the bracket when the bracket is written as x plus a number

How does the exponent affect the graph?+

For a nonzero scale a, even integer powers n ≥ 2 give line symmetry about x=h and a turning point at the centre. Odd integer powers n ≥ 3 give point symmetry about (h,k) and a stationary point of inflection

How are horizontal-axis intercepts found?+

Set y=0, isolate the power, and solve for every real branch allowed by the exponent and the isolated value

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 5: Higher-degree polynomials

In this chapter

Interpret and determine a transformed power functionConnect a factorised higher-degree rule and its graphSolve a higher-degree polynomial equation or inequalityAnalyse a parameterised polynomial familyBracket and approximate a polynomial root by bisectionApproximate solutions from intersections of two graphsLocate a polynomial turning point numerically