1. ## help on precalc. quadratic functions, ratios,

any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.

Put the quadratic function in standard form (i.e. .

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Given that , find the ratio .

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Use the properties of exponents to simplify the expression

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2. Originally Posted by rafaeli
any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.

Put the quadratic function in standard form (i.e. .

This is completing the square.

When in the form $\displaystyle ax^2 + bx + c$ where $\displaystyle a, \ b, \ c$ are constants; you can complete the square by $\displaystyle \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c$.

$\displaystyle x^2 + 5x+1$
$\displaystyle a=1, \ b=5, \ c=1$
$\displaystyle \left(x + \frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 + 1$
$\displaystyle \left(x + \frac{5}{2}\right)^2 - \frac{25}{4} + 1$
$\displaystyle \left(x + \frac{5}{2}\right)^2 - \frac{21}{4}$

Originally Posted by rafaeli
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Given that , find the ratio .
This question just involves algebra. Apply the rules provided and simplify.

$\displaystyle f(x+t) = \frac{x+t}{x+t+1}$

$\displaystyle f(x) = \frac{x}{x+1}$

$\displaystyle \implies \frac{f(x+t) - f(x)}{t} = \frac{\frac{x+t}{x+t+1} - \frac{x}{x+1}}{t}$

$\displaystyle \frac{\frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)}}{t}$

$\displaystyle \frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)(t)}$

$\displaystyle \frac{t}{(x+t+1)(x+1)(t)}$

$\displaystyle \frac{1}{(x+t+1)(x+1)}$

Originally Posted by rafaeli
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Use the properties of exponents to simplify the expression

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This question used the rules of indices.

$\displaystyle \left(\frac{9^{\frac14}}{2^{\frac12} 4^{\frac14}}\right)^{-2}$

$\displaystyle \left(\frac{9^{-\frac12}}{2^{-1} 4^{-\frac12}}\right)$

$\displaystyle \left(\frac{\frac13}{\left(\frac12\right)\left(\fr ac12\right)}\right)$

$\displaystyle \left(\frac{\left(\frac13\right)}{\left(\frac14\ri ght)}\right)$

$\displaystyle \left(\frac43\right)$

3. Originally Posted by rafaeli
any help on any of these is appreciated. i took precalc so long ago, i'm finding it really difficult to recall how to do certain things, and i didnt want to clutter the board with all these different questions.

Put the quadratic function in standard form (i.e. .

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Edit: Air, check your first step in completing the square. Should've added 1 instead of subtract. Just a typo, since the answer checked. I'm still too slow today,

Just complete the square.

$\displaystyle f(x)=(x^2+5x+ \ \ ?)+1 - \ \ ?$

Take half the coefficient of your linear term, square it and add it to make a perfect square trinomial, then subtract it to equalize the equation.

$\displaystyle f(x)=\left(x^2+5x+\frac{25}{4}\right)+1-\frac{25}{4}$

$\displaystyle f(x)=\left(x+\frac{5}{2}\right)^2+1-\frac{25}{4}$

$\displaystyle f(x)=\left(x+\frac{5}{2}\right)^2-\frac{21}{4}$

4. how do you get this?

Originally Posted by Air
This question just involves algebra. Apply the rules provided and simplify.

$\displaystyle f(x+t) = \frac{x+t}{x+t+1}$

$\displaystyle f(x) = \frac{x}{x+1}$

$\displaystyle \implies \frac{f(x+t) - f(x)}{t} = \frac{\frac{x+t}{x+t+1} - \frac{x}{x+1}}{t}$

$\displaystyle \frac{\frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)}}{t}$

$\displaystyle \frac{(x+t)(x+1) - (x)(x+t+1)}{(x+t+1)(x+1)(t)}$

$\displaystyle \frac{t}{(x+t+1)(x+1)(t)}$

$\displaystyle \frac{1}{(x+t+1)(x+1)}$

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5. Originally Posted by rafaeli
how do you get this?
You are given that $\displaystyle f(x) = \frac{x}{x+1}$ which is a given function and works for all value of $\displaystyle x$. If you are asked to work out the value of the function when $\displaystyle x=2$ (for example) when you would insert $\displaystyle 2$ into the $\displaystyle x$ places hence:

$\displaystyle f(x) = \frac{x}{x+1}$
$\displaystyle f(2) = \frac{2}{2+1} = \frac{2}{3}$

Similarly, when you are asked to work out $\displaystyle f(x+t)$, this means you are working out the value when $\displaystyle x = (x+t)$ hence inserting $\displaystyle x+t$ into $\displaystyle x$ places.

$\displaystyle f(x) = \frac{x}{x+1}$
$\displaystyle f((x+t)) = \frac{(x+t)}{(x+t)+1} = \frac{x+t}{x+t+1}$

6. thank you Air for clearing that up for me. you've been a great help.