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How do you find two quadratic function one that opens up and one that opens downward whose graphs have intercepts (0,0), (10,0)?

1 Answer

Jan 13, 2017

$f_{up} (x) = x^{2} - 10 x$ $f_"up"(x) = x^2-10x$

$f_{down} (x) = - x^{2} + 10 x$ $f_"down"(x) = -x^2+10x$

Explanation:

Any quadratic function with these $x$ $x$ intercepts is expressible in the form:

$f (x) = a x (x - 10) = a x^{2} - 10 a x$ $f(x) = ax(x-10) = ax^2-10ax" "$ for some constant $a \neq 0$ $a != 0$

It must take this form in order that $f (0) = 0$ $f(0) = 0$ and $f (10) = 0$ $f(10) = 0$ .

If $a > 0$ $a > 0$ then this will open upwards and if $a < 0$ $a < 0$ then it will open downwards.

So we can use $a = \pm 1$ $a=+-1$ to define the functions:

$f_{up} (x) = x^{2} - 10 x$ $f_"up"(x) = x^2-10x$

$f_{down} (x) = - x^{2} + 10 x$ $f_"down"(x) = -x^2+10x$

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