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3 years ago

9

What’s the standard form of the quadratic equation

1 answer:

Finger [1]3 years ago

6 0

y = ax^2 + bx + 7 because when x = 0 y = 7

when x = -1 y = 14 so we have
14 = a - b + 7
so a - b = 7...............(1)
when x = 1 y = 4
so 4 = a + b + 7
a + b = -3.................(2)
Adding equations (1) and (2) we eliminate b so:-
2a = 4
a = 2
Substitute a = 2 in equation (1)
2 - b = 7
b = -5

so our equation is y = 2x^2 - 5x + 7

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Answer:

$\boxed{\sf x=-2}$

$\boxed{\sf y=-4}$

Step-by-step explanation:

<u>First, Let's solve for x in -2x+2y=-4:</u>

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

<u>Subtract 2y from both sides:</u>

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

$\sf -2x=-4-2y$

<u>Divide both sides by -2:</u>

$\sf \cfrac{-2x}{-2}=-\cfrac{4}{-2}-\cfrac{2y}{-2}$

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<u>Now, we'll substitute x=y+2 to 3x+3y=-18:</u>

$\sf 3x+3y=-18$

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<u>Simplify:</u>

$\sf 6+6y=-18$

<u>Now, let's solve for y in 6+6y=-18</u>

$\sf 6+6y=-18$

<u>Subtract 6 from both sides:</u>

$\sf 6+6y-6=-18-6$

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<u>Divide both sides by 6:</u>

$\sf \cfrac{6y}{6}=\cfrac{-24}{6}$

$\bold{ y=-4}$

<u>Now, substitute y=-4 into x=2+y:</u>

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$\sf x=2+\bold{-4}$

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<u>_____________________________________</u>

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