How do you write a function rule for $x = 2, 4, 6$ $x = 2, 4, 6$ and $y = 1, 0, - 1$ $y = 1, 0, -1$ ?

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How do you write a function rule for $x = 2, 4, 6$ $x = 2, 4, 6$ and $y = 1, 0, - 1$ $y = 1, 0, -1$ ?

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Answer 1 · 2018-03-20T11:04:47

$y = f (x) = - \frac{1}{2} x + 2$ $y=f(x) = -1/2x+2$

Explanation:

Set the $i^{th}$ $i^("th")$ point as: $P_{i} \to (x_{i}, y_{i})$ $P_i->(x_i,y_i)$

Change in $x$ $x$ sequence is $2$ $2$
Change in $y$ $y$ sequence is $- 1$ $-1$

Gradient (slope) $\to m = \frac{change in y}{change in x} = - \frac{1}{2}$ $->m=("change in "y)/("change in " x) =-1/2$

Assuming there is a direct link between $x and y$ $x and y$ numbers of sequence we have:

$P_{1} \to (x_{1}, y_{1}) = (2, 1)$ $P_1->(x_1,y_1)=(2,1)$
$P_{2} \to (x_{2}, y_{2}) = (4, 0)$ $P_2->(x_2,y_2)=(4,0)$
$P_{3} \to (x_{3}, y_{3}) \to (6, - 1)$ $P_3->(x_3,y_3)->(6 ,-1 )$

Relating $m$ $m$ to, say, point 2 we have:

$m = - \frac{1}{2} = \frac{y - y_{2}}{x - x_{2}} = \frac{y - 0}{x - 4}$ $m=-1/2=(y-y_2)/(x-x_2)=(y-0)/(x-4)$

$- \frac{1}{2} = \frac{y - 0}{x - 4}$ $-1/2= (y-0)/(x-4)$

Multiply both sides by $(+ 2)$ $(+2)$

$- 1 = \frac{2 (y - 0)}{x - 4}$ $-1=(2(y-0))/(x-4)$

Multiply both sides by $(x - 4)$ $(x-4)$

$- (x - 4) = 2 y$ $-(x-4)=2y$

$- x + 4 = 2 y$ $-x+4=2y$

$y = - \frac{1}{2} x + 2$ $y=-1/2x+2$
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The question is specific in that it states 'function rule for $x$ $x$ '

So we write it as: $y = f (x) = - \frac{1}{2} x + 2$ $y=f(x) = -1/2x+2$

Tony B

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How do you write a function rule for $x = 2, 4, 6$ $x = 2, 4, 6$ and $y = 1, 0, - 1$ $y = 1, 0, -1$ ?

1 Answer

Explanation:

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How do you write a function rule for x=2,4,6x = 2, 4, 6 and y=1,0,−1y = 1, 0, -1?

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Explanation:

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How do you write a function rule for $x = 2, 4, 6$ $x = 2, 4, 6$ and $y = 1, 0, - 1$ $y = 1, 0, -1$ ?