Question

please help me
find the vertex of the parabola formed by the following quadratic function
y=4x^2+8x+7

Answer 1

given the form:

y = ax^2 + bx + c

the vertex can be determined as:

x = -b/(2a)
y = (4ac-b^2)/(4a)

Answer 2

don't get it
so is it (-2,7)

Answer 3

you might want to better define the parts to use those formulas

Answer 4

so how do I do y=2x^2-8x+2
using the vertex of the parabola formed by the following quadtic function

Answer 5

define the abc parts

Answer 6

a=-8x
b=2x^2
c=+2
I think

Answer 7

lets refine that a little bit

ax^2 + bx + c
2x^2 -8x +2

a = 2
b = -8
c = 2

Answer 8

so now how do I find the quadratic function

Answer 9

\[x=-\frac b{2a}=-\frac{-8}{2(2)}\]
\[y=\frac {4ac-b^2}{4a}=\frac {4(2)(2)-8^2}{4(2)}\]

Answer 10

if we want we can prolly simplify this alot by saying:
\[y=c-\frac{b^2}{4a}=2-\frac{8^2}{4(2)}\]

Answer 11

so do I get the answer

Answer 12

those are the answers ...

Answer 13

but its out of (-4,66) or(4,2) or (-2,26) or (2,-6)

Answer 14

you have to do some of the work, ive reduced it down to basic math for you. I am not going to add and multiply for you as well

Answer 15

so I multiply 4 (2)

Answer 16

\[x=-\frac{-8}{2(2)}\]
\[y = 2-\frac{8^2}{4(2)}\]

Answer 17

and yes 4(2) means multiplication

Answer 18

i would suggest that since they all have different x values, that you only need to determine the value of x and not y

Answer 19

so the 2 (2)

Answer 20

\[y=4x ^{2}+8x+7=4\left( x ^{2}+2x \right)+7=4\left( x ^{2}+2x+1-1 \right)+7\]
\[y=4\left( x+1 \right)^{2}-4+7\]
\[y-3=4\left( x+1 \right)^{2}\]
vertex is (-1,3)