<u>Answer:</u>
<u>Step-by-step explanation:</u>
We have a parabola and a small line shown in the graph.
The parabola goes upto x = 2 but does not reach that very value. So x = 2 starts on the line above where x = 2 and ends just before x = 4.
Therefore, this function can be modeled by:
Answer:
Vertex= (-5,-4)
Another point=(0, -129)
Step-by-step explanation:
The vertex is also the maximum, and the point is also the y-intercept.
This looks complicated
lets simplify both sides
By multiplying top and bottom by the conjugate:-
left side = (5-3i)(x + iy) (4+5i) (5-3i)(x + iy) (4+5i)
------------------------ = ------------------------
(4-5i)(4+5i) 41
right side = (2 + i + 3-4)(2+i - 3 + 4i) (by Difference of 2 squares)
= 5-3i)(-1 + 4i)
so as left side = right side
(5-3i)(x+yi)(4+5i) = 41 (5-3i)(-1+5i)
5-3i is common so:-
(x + iy)(4+5i) = 41(-1+5i)
4x +5xi + 4yi - 5y = -41 + 205i
4x - 5y + (5x + 4y)i = -41 + 205i
EQuating coefficients we are left with the system of equations
4x - 5y = -41
5x + 4y = 205
solving this gives x = 21 and y = 25
You need a length and width for it
