Question

can someone PLEASE please please pleeease help me! I'll reward you with a fan and medal.

which of the following has the greatest x value for the vertex? explain your choice.

A. f(x) = x^2 +4x + 4
B. f(x) = 3(x-1)^2+5
C. f(x) = -2(x+2)^2+3
D. f(x) = -3x^2+6x+1

Answer 1

A. f(x)=x^2+4x+4=(x+2)^2 vertex is (-2,0)
B.f(x)=3(x-1)^2+5 vertex is (1,5)
C.f(x)=-2(x+2)^2+3 vertex is (-2,3)
D.f(x) = -3x^2+6x+1=-3(x^2-2x-1)=-3(x-1)^2+6 vertex is (1,6)

Answer 2

@rosamartinez

Answer 3

so, A? @jango_IN_DTOWN

Answer 4

@Michele_Laino

Answer 5

the x-coordinate of the vertex of a generic parabola:
\(y=ax^2+bx+c\)
is:
\[x = - \frac{b}{{2a}}\]

Answer 6

I have no idea how to do this, honestly.

Answer 7

for example, if we consider the parabola in A) we get:
\[x = - \frac{b}{{2a}} = - \frac{4}{{2 \cdot 1}} = - 2\]
being: \(a=1, b=4\)

Answer 8

now, I consider the parabola in B).
I can rewrite such parabola as below:
\[\begin{gathered}
y = 3{\left( {x - 1} \right)^2} + 5 = 3\left( {{x^2} + 1 - 2x} \right) + 5 = \hfill \\
\hfill \\
= 3{x^2} + 3 - 6x + 5 = 3{x^2} - 6x + 8 \hfill \\
\end{gathered} \]

Answer 9

I have computed the square of the binomial: \((x-1)^2\)
so we get:
\[y = 3{x^2} - 6x + 8\]
as we can see, by comparison with the generic formula of a parabola, we have:
\(a=3,b=-6,c=8\)
so, using my formula above, we get:
\[x = - \frac{b}{{2a}} = - \frac{{ - 6}}{{2 \cdot 3}} = 1\]
namely the vertex of that parabola is located at \(x=1\)

Answer 10

please do the same with the parabola D.

Answer 11

hint:
there we have: \(a=-3,b=6,c=1\)
so:
\[x = - \frac{b}{{2a}} = - \frac{6}{{2 \cdot \left( { - 3} \right)}} = ...?\]

Answer 12

please complete

Answer 13

ok!

Answer 14

it is simple:
here is the answer:
\[x = - \frac{b}{{2a}} = - \frac{6}{{2 \cdot \left( { - 3} \right)}} = 1\]

Answer 15

so is it A, B, C or D?

give an example of a perfect square trinomial and the difference of squares binomial, the show how to factor.

144r^2 + 24r + 1
5 5^2 25
A difference of squares can be factored using the following pattern: a2−b2=(a+b)(a−b)


@Michele_Laino

Answer 16

an example of the perfect square trinomial, can be this:
\(x^2+49+14x=(x+7)^2\)

Answer 17

the pattern is:
\((a+b)^2=a^2+b^2+2 a b\)

Answer 18

for example, if we have to compute the square of this binomial:
\((12r+1)^2\)
we have \(a=12r, b=1\)
so, if we apply the pattern above, we can write:
\(a^2+b^2+2 a b = (12r)^2+ (1)^2+ 2 \cdot 12 r \cdot 1=...\)
please complete

Answer 19

hint:
\((12r)^2=12r \cdot 12r= (12 \cdot 12) r^2=...?\)

Answer 20

by the way, for first question, the answer is B. and D.

Answer 21

what is \( 12 \cdot 12=...?\)

Answer 22

144

Answer 23

correct!

Answer 24

so we have:
\[{\left( {12r + 1} \right)^2} = 144{r^2} + 1 + 24r\]
furthermore, a difference of square binomial, can be this:
\(25x^2-49\)=\((5x-7)\cdot(5x+7)\)

Answer 25

okay, so the first one is a square trinomial and the second a square binomial right? and for the first question it can only be one choice and I need to explain my choice. @Michele_Laino

Answer 26

more precisely, the first one:
\((12r+1)^2=144r^2+1+24r\) is the square trinomial
whereas the second one: \(25x^2-49=(5x-7)(5x+7)\) is the difference of squares

Answer 27

the vertex for each parabola is:
\((-2,0)\) for parabola in A
\((1,5)\) for parabola in B
\((-2,3)\) for parabola in C
\((1,4)\) for parabola in D

Answer 28

if we are looking for the vertex with the highest y-coordinate, then our answer is option B