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AI SL · 12.5

Functions

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12.5.2

Key features and models of graphs

12.5.2

Key features and models of graphs

Target: I can find the intercepts, zeros, vertex and axis of symmetry of a quadratic model, and say what they mean. (+4 more)

By the end of this section

Key points

Key features of a quadratic model

  • \(f(x)=ax^2+bx+c\) has its y-intercept at \((0,\,c)\). Its zeros are the roots of \(ax^2+bx+c=0\).
  • The vertex is the maximum (\(a<0\)) or minimum (\(a>0\)). The axis of symmetry is the vertical line through the vertex.
  • Use the calculator for the vertex and the zeros. In a context, reject a zero that makes no sense (a negative distance).

Exponential models and asymptotes

  • \(f(x)=ka^x+c\) and \(f(x)=ke^{rx}+c\) have the horizontal asymptote \(y=c\).
  • In a cooling model the asymptote is the surrounding temperature. In a growth or decay model it is the limit the quantity approaches.
  • The constant c is not the starting value: the starting value is \(f(0)=k+c\).

Direct and inverse variation

  • \(y=ax^n\) is direct variation when n is positive: \(y\propto x^n\). Find a by putting in one known pair of values.
  • \(y=\dfrac{a}{x^n}\) (negative power) is inverse variation. The y-axis is a vertical asymptote and the x-axis is a horizontal asymptote.
  • Say in words what the asymptote means: very small x gives a very large y.

Rational functions and choosing a model

  • For \(f(x)=\dfrac{ax+b}{cx+d}\): the vertical asymptote is \(x=-\frac{d}{c}\) and the horizontal asymptote is \(y=\frac{a}{c}\).
  • The x-intercept is where \(ax+b=0\). The y-intercept is \(f(0)=\frac{b}{d}\).
  • Constant differences mean a linear model. Constant ratios mean an exponential model. A rise then a fall (or a fall then a rise) suggests a quadratic.

On the calculator

Finding a vertex and zeros

  • Graph the quadratic and choose a window that shows the turning point and the places where it crosses the x-axis.
  • Use the maximum (or minimum) tool for the vertex, and the zero tool for each x-intercept.
  • Read the y-intercept from the table at \(x=0\).

Screenshot and video to come

Solving a model for a time

  • Enter the model as Y1 and the target value as Y2.
  • Choose a window that shows where they meet and use the intersection tool.
  • The x-coordinate is the time. Round as the question asks.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
The top of a jet of water

A fountain sends water upwards and outwards. At a horizontal distance of x metres from the nozzle the jet is at height \(h(x)=-0.5x^2+3x+0.4\) metres. Find the greatest height of the jet and the horizontal distance at which the water reaches the ground. Give your answers to 3 significant figures.

Working
  1. Graph the model

    Enter \(h(x)\) as Y1. Choose a window with x from 0 to 8.

    The graph is a hill: the water rises, then falls.

  2. The vertex

    From the calculator: maximum at \((3,\,4.9)\)

    The vertex gives the greatest height, 4.9 m, at 3 m from the nozzle. The axis of symmetry is \(x=3\).

  3. The zero

    From the calculator: zero at \(x=6.13\)

    The water lands where \(h=0\). The other zero, \(x=-0.130\), is behind the nozzle, so it is rejected.

  4. The y-intercept

    \(h(0)=0.4\)

    The nozzle is 0.4 m above the ground.

Video walkthrough coming
A cooling drink

The temperature of a cup of coffee, in °C, is modelled by \(T(t)=20+75e^{-0.04t}\), where t is the time in minutes after it is poured. Find the temperature the coffee approaches, and the time it takes to cool to 30 °C. Give the time to 3 significant figures.

102030405060708020406080100t (minutes)T (°C)T
Working
  1. The start

    \(T(0)=20+75=95\)

    At the start the exponential term is at its largest.

  2. The long run

    As t grows, \(e^{-0.04t}\to0\), so \(T\to20\). The horizontal asymptote is \(T=20\).

    The coffee cools towards room temperature and never goes below it.

  3. Solve for t

    \(20+75e^{-0.04t}=30\). From the calculator: \(t=50.4\)

    Graph \(Y1=T(t)\) and \(Y2=30\) and find the intersection, or solve for t.

Video walkthrough coming
Inverse variation

The time t hours that a team takes to paint a fence is inversely proportional to the number of painters n, so \(t=\dfrac{a}{n}\). Four painters take 9 hours. Find a, and find the time for 6 painters.

Working
  1. Find a

    \(9=\dfrac{a}{4} \;\Rightarrow\; a=36\)

    Put the known pair \((4,\,9)\) into the model.

  2. The model

    \(t=\dfrac{36}{n}\)

    Now any number of painters can be put in.

  3. Predict

    \(t=\dfrac{36}{6}=6\)

    6 painters take 6 hours.

  4. The asymptote

    As n gets close to 0, t grows without limit: the t-axis (\(n=0\)) is a vertical asymptote.

    With very few painters the job takes very long. The graph never touches the t-axis.

Video walkthrough coming
Intercepts and asymptotes

The function f is defined by \(f(x)=\dfrac{6x-9}{2x+4}\), \(x\neq-2\). Write down the equations of the asymptotes and the coordinates of the intercepts with the axes.

−8−6−4−22468−4−2246810xy
Working
  1. Vertical asymptote

    \(2x+4=0 \;\Rightarrow\; x=-2\)

    The function is undefined where the denominator is 0.

  2. Horizontal asymptote

    \(y=\dfrac{6}{2}=3\)

    For large x only the x terms matter: \(\frac{6x}{2x}\).

  3. The x-intercept

    \(6x-9=0 \;\Rightarrow\; x=1.5\)

    A fraction is 0 when its top is 0.

  4. The y-intercept

    \(f(0)=\dfrac{-9}{4}=-2.25\)

    Put \(x=0\) into the function.

Video walkthrough coming
Which model fits?

A laptop is bought for £1200. Its value at the end of each year is shown in the table. Decide whether a linear, quadratic or exponential model fits, give a reason, and write the model.

Year, t0123Value, £1200960768614.40
Working
  1. Differences

    \(960-1200=-240,\; 768-960=-192\)

    The differences are not constant, so the model is not linear.

  2. Ratios

    \(\dfrac{960}{1200}=0.8,\; \dfrac{768}{960}=0.8,\; \dfrac{614.4}{768}=0.8\)

    The ratio is constant: the laptop keeps 80% of its value each year.

  3. The model

    \(V(t)=1200\times0.8^{t}\)

    Exponential decay. The value approaches 0 but never reaches it.

Video walkthrough coming