Search topics

Search

Analyse and optimise a quadratic model

Start a set of 8
Step 3 of 6

Analyse a quadratic model

A quadratic model only describes the situation for inputs in its meaningful domain. The turning point gives the optimum when its input sits inside the domain; otherwise the optimum is at an endpoint, and threshold roots outside the domain are discarded

\[y=a(x-h)^2+k,\quad x_{\min}\le x\le x_{\max},\qquad h(t)=-(t-5)^2+30\]
  1. 1State the meaningful domain Use only inputs in the interval for which the model describes the situation \(\displaystyle 0\le t\le 4\)
  2. 2Find the turning point Locate the turning point input, then evaluate the model there to find the algebraic optimum \(\displaystyle t_v=5,\quad h(5)=30\)
  3. 3Use the coefficient sign A negative leading coefficient gives a maximum at the turning point, while a positive one gives a minimum \(\displaystyle a=-1<0\Rightarrow\text{maximum}\)
  4. 4Solve a threshold equation Set the model output equal to the stated level and keep only inputs in the meaningful domain \(\displaystyle a(x-h)^2+k=L\Rightarrow(x-h)^2=\frac{L-k}{a}\)
Year 11, VCE Units 1 & 2, Chapter 3, 4 difficulty levels, Calculator allowedBuild a worksheet

Analyse a quadratic model

Use the turning point of a quadratic model to identify an algebraic maximum or minimum. Check the model domain before interpreting the turning point, and solve level equations only for inputs that the model allows

\[y=a(x-h)^2+k,\quad x_{\min}\le x\le x_{\max}\]
  1. State the meaningful domain Use only inputs in the interval for which the model describes the situation \(\displaystyle x_{\min}\le x\le x_{\max}\)
  2. Find the turning point For a general quadratic, calculate the input coordinate and then evaluate the model there \(\displaystyle x_v=-\frac{b}{2a},\quad y_v=f(x_v)\)
  3. Use the coefficient sign A negative leading coefficient gives a maximum at the turning point, while a positive one gives a minimum \(\displaystyle a<0\Rightarrow\text{maximum},\quad a>0\Rightarrow\text{minimum}\)
  4. Solve a threshold equation Set the model output equal to the stated level and keep only inputs in the meaningful domain \(\displaystyle a(x-h)^2+k=L\Rightarrow(x-h)^2=\frac{L-k}{a}\)

Worked example

The height model and domain are shown below, with time in seconds and height in metres. Find the maximum height and the times when the height is 14 metres

\[h(t)=-t^2+8t+2,\quad0\le t\le7,\quad h=14\]
  1. \(\displaystyle t_v=-\frac{8}{2(-1)}=4\)The turning-point time is found from the coefficients of the general-form rule
  2. \(\displaystyle h(4)=-(4)^2+8(4)+2=18\)Substitute the turning-point time into the model to find the maximum height
  3. \(\displaystyle -t^2+8t+2=14\Rightarrow-t^2+8t-12=0\Rightarrow t^2-8t+12=0\)Set the model equal to 14 and rearrange the equation to zero form
  4. \(\displaystyle t^2-8t+12=(t-2)(t-6)=0\Rightarrow t=2\text{ or }t=6\)Both threshold times are in the domain from 0 to 7 seconds. The maximum height is 18 metres at 4 seconds; the height is 14 metres at 2 seconds and 6 seconds

Common mistakes

Using the unrestricted vertex
✕\(\displaystyle q(t)=-(t-7)^2+50,\quad0\le t\le5\Rightarrow\text{maximum at }(7,50)\)✓\(\displaystyle q(0)=1,\quad q(5)=46\Rightarrow\text{maximum at }(5,46)\)
The vertex at t = 7 is outside the stated domain, so the feasible endpoints must be compared
Swapping vertex coordinates
✕\(\displaystyle h(t)=-(t-4)^2+16\Rightarrow\text{maximum height}=4\)✓\(\displaystyle h(t)=-(t-4)^2+16\Rightarrow\text{maximum height}=16\text{ at }t=4\)
The first coordinate is the input where the optimum occurs, while the second coordinate is the output value
Keeping an infeasible crossing
✕\(\displaystyle h(t)=-(t-4)^2+16,\quad0\le t\le5,\quad h=7\Rightarrow t=1\text{ or }7\)✓\(\displaystyle h(t)=-(t-4)^2+16,\quad0\le t\le5,\quad h=7\Rightarrow t=1\)
A threshold input is usable only when it lies in the model's stated domain

Difficulty levels

introRead an integral interior vertex and its output for a model with a closed domain
standardUse exact features, a domain check and one threshold input for a quadratic model
challengeHandle an endpoint optimum or two threshold crossings, including exact irrational inputs
examAnalyse a model, interpret its domain and threshold results, and round a final output when requested

Questions about this topic

How is the turning-point input found from a general quadratic?+

For a quadratic in general form, calculate the turning-point input using the formula below, then substitute it into the rule to find the output coordinate

\[y=ax^2+bx+c,\qquad x=-\frac{b}{2a}\]
Why must the domain be checked?+

The algebraic rule may have a turning point outside the situation being modelled. If that happens, compare the feasible endpoints instead

How are threshold inputs found?+

Set the quadratic output equal to the stated level, solve the resulting equation and keep only roots in the meaningful domain

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 3: Quadratic functions

In this chapter

Factorise, solve and reconstruct a rational quadraticCalculate and interpret a quadratic discriminantTransform a quadratic and find its key featuresComplete the square and link equivalent formsDetermine a quadratic rule from exact informationSolve a quadratic exactly with the quadratic formulaConstruct a rational quadratic from a surd rootSolve an equation that reduces to quadratic formAnalyse signs and solve a quadratic inequalityIntersections of a line and a quadratic graphDetermine parameter values from a discriminant conditionFormulate and solve a quadratic problemAnalyse and optimise a quadratic model