# Function Graphs

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## Function Graphs

You have so far seen graphs of straight lines, quadratics and cubics. Now it is time to see some graphs of more complicated functions, such as graphs of $y = kx^n$ and quartics. As before, you will only need to draw a rough sketch of these graphs.

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## Graphs of $y = kx^n$

Graphs of $y = kx^n$ have the same basic shape, however the values of $k$ and $n$ affect how they look.

Here are some examples:

$n$ positive and even

The graphs are either:

• u-shaped if $k$ is positive
• n-shaped if $k$ is negative

$n$ positive and odd

The graphs either go:

• bottom-left to top-right if $k$ is positive
• top-left to bottom-right if $k$ is negative

(the graphs go ‘corner-to-corner’)

$n$ negative and even

The graphs are mirror images in the $y$-axis, and are:

• above the $x$-axis if $k$ is positive
• below the $x$-axis if $k$ is negative

$n$ negative and odd

The parts of the graph are in diagonally opposite quadrants, and are:

• in the bottom-left and top-right quadrants if $k$ is positive
• in the top-left and bottom-right quadrants if $k$ is negative

Note: an asymptote is a line that the curve gets infinitely close to, but never touches. The third and fourth graphs both have asymptotes at $x=0$ and $y=0$.

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## Quartics

A quartic is a polynomial with an $x^4$ term as the highest power. To sketch a quartic, you will need to find where the curve crosses or touches the $x$-axis – the expression will be usually factorised which will make it easier to find these values.

Quartics with positive $x^4$ coefficients are positive for very positive and very negative $x$-values. Quartics with negative $x^4$ coefficients are negative for very positive and very negative $x$-values.

Example: Sketch the graph of $f(x) = x^2(x-1)(x+2)$

Let $f(x)=0$ to find the points where the curve crosses the $x$-axis:

$x^2(x-1)(x+2) = 0$

$x^2$ is a double root, so the graph touches the $x$-axis at $\textcolor{red}{0}$

The curve crosses the $x$-axis at $\textcolor{red}{1}$ and $\textcolor{red}{-2}$

Substitute in $x=0$ to find where the curve crosses the $y$-axis:

$y = 0^2(0-1)(0-2) = \textcolor{red}{0}$

The coefficient of $x^4$ is positive, so the curve will be positive for very positive and very negative $x$-values.

Hence, we have enough information to sketch the graph.

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## Function Graphs Example Questions

Question 1: Sketch the graph of $y=-2x^5$, labelling any points of intersection with the axes.

[2 marks]

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The value of $n$ is positive and odd $(5)$, and the value of $k$ is negative $(-2)$.

Hence, the graph will have a corner-to-corner shape from top-left to bottom-right.

The graph will pass through the origin.

Therefore, the graph will look like this:

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Question 2: Sketch the graph of $y=\dfrac{4}{x^{3}}$, labelling any points of intersection with the axes.

[2 marks]

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$y=\dfrac{4}{x^{3}} = 4x^{-3}$

The value of $n$ is negative and odd $(-3)$, and the value of $k$ is positive $(4)$.

Hence, the graph will be in the bottom-left and top-right corner.

Therefore, the graph will look like this:

Gold Standard Education

Question 3: Sketch the graph of $y=-(x-1)^2 (x+1)^2$, labelling any points of intersection with the axes.

[3 marks]

A Level AQAEdexcelOCR

The coefficient of $x^4$ is negative, therefore the curve will be negative for very positive and very negative $x$-values.

Find the points where the curve crosses the $x$-axis:

$-(x-1)^2 (x+1)^2 = 0$

$(x-1)^2$ and $(x+1)^2$ are both double roots, therefore the curve will touch the $x$-axis at $x=1$ and $x=-1$

Find the point where the curve intersects the $y$-axis:

$y = -(0-1)^2 (0+1)^2 = -1$

Therefore, we have enough information to sketch the graph:

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