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3 years ago

Which order pair Is a solution of y= -6x - 7

1 answer:

6 0

(x,y)
so basically an easy way to find is to subsitute values for x nd get values for y

if x=0 then
y=-6(0)-7
y=0-7
y=-7
one solution is (0,-7)

another
if x=1
y=-6(1)-7
y=-6-7
y=-13
another is (1,-13)

there are many
just try inputing the first number iin for x and see if the second number equals y
pleas include the ordered pairs

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First off, we factor out the expression:

$\displaystyle \large{y = 2 {x}^{2} - 12x + 16} \\ \displaystyle \large{y = 2 ( {x}^{2} - 6x + 8) }$

In the bracket, separate 8 out of the expression.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 8)] }\\ \displaystyle \large{y = 2[ ( {x}^{2} - 6x) + 8]}$

In x^2-6x, find the third term that can make up or convert it to a perfect square form. The third term is 9 because:

$\displaystyle \large{ {(x - 3)}^{2} = {x}^{2} - 6x + 9}$

So we add +9 in x^2-6x.

$\displaystyle \large{y = 2[ ( {x}^{2} - 6x + 9) + 8]}$

Convert the expression in the small bracket to perfect square.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8]}$

Since we add +9 in the small bracket, we have to subtract 8 with 9 as well.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} + 8 - 9]} \\ \displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]}$

Then we distribute 2 in.

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\$

$\displaystyle \large{y = 2[ {(x - 3)}^{2} - 1]} \\ \displaystyle \large{y = [2 \times {(x - 3)}^{2} ]+[ 2 \times ( - 1)] } \\ \displaystyle \large{y = 2 {(x - 3)}^{2} - 2 }$

Remember that negative multiply positive = negative.

Hence the vertex form is y = 2(x-3)^2-2 or first choice.

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