Y=-x²+6x-11; y=-6

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вопрос

To find the point of intersection between the two equations, we need to set them equal to each other:

-x^2 + 6x - 11 = -6

Now, let's simplify the equation:

-x^2 + 6x - 11 + 6 = 0
-x^2 + 6x - 5 = 0

Now to solve for x, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = -1, b = 6, and c = -5. Plugging these values into the formula:

x = (-(6) ± sqrt((6)^2 - 4(-1)(-5))) / 2(-1)
x = (-6 ± sqrt(36 - 20)) / -2
x = (-6 ± sqrt(16)) / -2
x = (-6 ± 4) / -2

Now we have two possible values for x, which are:

x = (-6 + 4) / -2 = -2 / -2 = 1
x = (-6 - 4) / -2 = -10 / -2 = 5

So the points of intersection are when x = 1 and x = 5. To find the corresponding y-values, we can plug them back into one of the original equations. Let's use the first equation:

y = -1^2 + 6(1) - 11
y = -1 + 6 - 11
y = -6

Therefore, the point of intersection is at (1, -6).