MHB Finding the vertex of a quadratic and the product of two complex numbers

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The discussion focuses on finding the vertex of the quadratic function f(x) = -2x^2 - 8x + 3 and multiplying two complex numbers. The initial answer provided for the vertex was (-2, 0), but it was corrected to (-2, 11) after further calculations. The correct method involved using the vertex formula x = -b/(2a) and substituting back to find the y-coordinate. The multiplication of the complex numbers (6 - 5i)(4 + 3i) was confirmed to yield the correct answer of 39 - 2i. The conversation emphasizes the importance of verifying calculations in algebraic problems.
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Basically I don't know anyone in real life that can help me with this, so I need help checking to see if my answers are correct :)

PART A

11) Find the vertex of f(x) = -2x^2 - 8x + 3 algebraically.

My Answer: (-2,0)

12) Multiply and simplify: (6 - 5i) (4 + 3i)

My Answer: 39 - 2i
 
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Re: Please check my answers - 6

drop said:
Basically I don't know anyone in real life that can help me with this, so I need help checking to see if my answers are correct :)

PART A

11) Find the vertex of f(x) = -2x^2 - 8x + 3 algebraically.

My Answer: (-2,0)

I get $(-2, 11)$. Now can you help me by explaining to us how did you arrive at $(-2, 0)$ as the vertex of the function of f of x?
drop said:
12) Multiply and simplify: (6 - 5i) (4 + 3i)

My Answer: 39 - 2i

Correct.:)
 
Re: Please check my answers - 6

x = -b/(2a)

x = -(-8)/2(-2)

x = 8/-4 = -2

I think I forgot to substitute -2 for x to find out y, thanks for pointing that out! (I was thinking of x intercepts) After solving for y, I got (-2,11) Thank you
 
Re: Please check my answers - 6

Yes...and as you can see, $f(-2)=-2(-2)^2-8(-2)+3=11$.
 

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