Question #705e3

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Question #705e3

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Answer 1 · 2017-07-08T18:19:44

Factorising

Explanation:

If you factorise this polynomial, you will get:
$f (x) = x (x^{2} - 9)$ $f(x)=x(x^2-9)$

The quadratic is a special one because it is the difference of two squares i.e. a square number minus another square number. When you spot this, you can factorise it further:

$(x^{2} - 9) = (x + 3) (x - 3)$ $(x^2-9)=(x+3)(x-3)$

giving you

$f (x) = x (x + 3) (x - 3)$ $f(x)=x(x+3)(x-3)$

These are the three points where the graph crosses the x axis (0, -3 and 3 respectively) and if you do f(0) (or put x = 0) that gives you the y intercept which is 0 in this case.

Also remember that positive cubics have a kind of capital N shape when sketching (but it's curved :) )

Answer 2 · 2017-07-08T18:28:45

$see explanation$ $"see explanation"$

Explanation:

$since this is in the geometry section I will not use calculus$ $"since this is in the geometry section I will not use calculus"$

$find the x-intercepts (roots ) by equating to zero$ $"find the x-intercepts (roots ) by equating to zero"$

$\Rightarrow x^{3} - 9 x = 0 \leftarrow now factorise$ $rArrx^3-9x=0larr" now factorise"$

$\Rightarrow x (x^{2} - 9) = 0 \leftarrow x^{2} - 9 is difference of squares$ $rArrx(x^2-9)=0larr x^2-9" is difference of squares"$

$\Rightarrow x (x - 3) (x + 3) = 0$ $rArrx(x-3)(x+3)=0$

$equate each factor to zero$ $"equate each factor to zero"$

$x = 0 \Rightarrow x = 0$ $x=0rArrx=0$

$x - 3 = 0 \Rightarrow x = 3$ $x-3=0rArrx=3$

$x + 3 = 0 \Rightarrow x = - 3$ $x+3=0rArrx=-3$

$since polynomial is of degree 3 (highest power of x )$ $"since polynomial is of degree 3 (highest power of x ) "$
$and has a positive leading coefficient$ $"and has a positive leading coefficient "$

$then graph starts down and ends up$ $"then graph starts down and ends up"$

$we can choose values of x between the roots as an$ $"we can choose values of x between the roots as an "$
$indication of the shape of the graph$ $"indication of the shape of the graph"$

$f (- 1) = - 1 + 9 = 8 \leftarrow above x-axis$ $f(-1)=-1+9=8larrcolor(red)" above x-axis"$

$f (1) = 1 - 9 = - 8 \leftarrow below x-axis$ $f(1)=1-9=-8larrcolor(red)" below x-axis"$
graph{x^3-9x [-20, 20, -10, 10]}

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Question #705e3

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