Question

What is the vertex of the graph of y = −3x^2 − x + 8?

Answer 1

Step by step please? x :)

Answer 2

step by step:
the first coordinate of the vertex of \(y=ax^2+bx+c\) is \(-\frac{b}{2a}\)
in your example \(a=-3,b=-1\)

Answer 3

so what would you get for \(-\frac{b}{2a}\) if \(a=-3,b=-1\) ?

Answer 4

-1/6

Answer 5

ok good
now for the second coordinate, replace \(x\) in
\[y=-3x^2-x+8\] by \(-\frac{1}{6}\) and find \(y\)
i.e. compute
\[-3(-\frac{1}{6})^2-(-\frac{1}{6})+8\]

Answer 6

the arithmetic is kind of annoying, but totally doable

Answer 7

I think thats my problem, I'm not multiplying correctly :\

Answer 8

\(-3\times \frac{1}{36}+\frac{1}{6}+8\)
\(=-\frac{1}{12}+\frac{1}{6}+8\)
\(=-\frac{1}{12}+\frac{2}{12}+8\)
\(=\frac{1}{12}+8\)

Answer 9

so either \(8\tfrac{1}{12}\) or \(\frac{97}{12}\)

Answer 10

seem reasonable?

Answer 11

The -1/12 part is whats kind of confusing me. It's probably dumb of me but how did you get it? I did everything else correctly

Answer 12

ok so that part comes from
\[-3\times \left(-\frac{1}{6}\right)^2\]

Answer 13

square first, multiply second
\[(-\frac{1}{6})\times( -\frac{1}{6})=\frac{1}{36}\]

Answer 14

then \(-3\times \left(\frac{1}{36}\right)=\frac{-3}{36}\)

Answer 15

reduce this fraction by dividing top and bottom by 3 and get \(-\frac{1}{12}\)

Answer 16

hows that?

Answer 17

That makes more sense, can we try another?

Answer 18

sure i'm ready

Answer 19

Where does the graph of y = 6x^2 − 11x − 10 cross the x-axis? I'm kinda rusty on it... -b/2a then plug in the numbers right?

Answer 20

this is completely different because you are not asked for the vertex, you are asked for where it crosses the \(x\) axis
it will cross the \(x\) axis if \(y=0\) so your job here is not to find the vertex, but rather to solve
\[6x^2-11x-10=0\] for \(x\)

Answer 21

fortunately this one factors as
\[(2x-5)(3x+2)=0\]

Answer 22

if you don't like factoring you can use the quadratic formula, but this one was cooked up to factor
do you know how to solve
\[(2x-5)(3x+2)=0\] for \(x\) ?

Answer 23

Can you just keep going so I can see all of your steps, I get it but then again I don't so please continue :)

Answer 24

ok so first you do understand i hope that you are not looking for the vertex, but rather you are looking to find the values of \(x\) for which
\[6x^2-11x-10=0\]
since this factors as
\[(2x-5)(3x+2)=0\] we can set each factor equal to zero and solve

Answer 25

if \(2x-5=0\) then \(2x=5\) and so \(x=\frac{5}{2}\) is one solution

Answer 26

similarly solve
\[3x+2=0\\3x=-2\\x=-\frac{2}{3}\]

Answer 27

so it crosses the \(x\) axis at two point, \((-\frac{2}{3},0)\) and \((\frac{5}{2},0)\)

Answer 28

this ok?

Answer 29

Yeah, I'm going to study this. I'll be back in a few minutes, thank you :)

Answer 30

yw

Answer 31

Part 1 − Find the vertex, axis of symmetry, domain, and range of the graph of y = −3x2 − 3x + 4. Show all work for full credit.

Part 2 − Using complete sentences, explain how you can determine the axis of symmetry, the domain, and range without graphing y = −3x2 − 3x + 4.


How do you find the domain & range

Answer 32

did you repost this question?