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## Question:

Which equation could generate the curve in the graph below? y = –2×2 + 3x – 5 y = –2×2 – 4x – 2 y = –2×2 – 16x– 28 y = –2×2 +16x –28

## Answer:

In order to find which equation could generate the curve, we can take each option and verify if delta if greater than zero because we have 2 intersection points with OX and if those intersection points are both negative (the intersection point are the solution of the equation).

First option: delta = 3^2-4*(-2)*(-5) = 9-40<0 not a good option

Second option: delta = 4^2-4*(-2)*(-2) = 16-16=0<0 not a good option

Third option: delta = 16^2-4*(-2)*(-28) = 256-224 = 32

x1,2 = (16+-√32)/(-4) = -4-+√2

Both values are negative and delta<0 so this is a good solution.

Fourth option: delta = 16^2-4*(-2)*(-28) = 256-224 = 32

x1,2 = (-16+-√32)/(-4) = 4-+√2. Just one solution is negative the other one is positive. Not a good solution.

The final equation is:

y = –2x^2 – 16x– 28

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