Question

Find a and b. The parabola y=ax^2+bx-3 has vertex (3,15).

Answer 1

Have you tried using the vertex formula for your equation y=ax^2+bx+c
(-b/(2a), f(-b/(2a))

Answer 2

\[(3,15)=(\frac{-b}{2a},f(\frac{-b}{2a}))\]

Answer 3

yes but I keep getting it wrong therefore would you please help with the answer?

Answer 4

yep. just wondering if you had tried it.

Answer 5

so we have two equations:
\[3=\frac{-b}{2a} \]



\[15=f(\frac{-b}{2a})\]

Answer 6

\[3=\frac{-b}{2a} => 6a=-b => -6a=b\]
Go back to your y=f(x) equation and do f(3) and replace the b with -6a.

Answer 7

Show me what you have when you do this.

Answer 8

\[f(x)=ax^2+bx-3\]
Replace the x's with 3 and the b with (-6a)
and then remember we can say f(-b/(2a)) or f(3)=15 since -b/(2a) is 3.

Answer 9

so a= 2 and b= 3>

Answer 10

that wouldn't work. 3 does not equal -3/(2*2)

Did you try what I said? I was hoping you would just perform the one step I asked you to do.

Answer 11

a = -3 and b = -15?

Answer 12

We have \[f(x)=ax^2+bx-3 \]
I asked you to replace the x's with 3's like this:
\[f(3)=a(3)^2+b(3)-3\]
So this means after simplifying we have
\[f(3)=a(9)+b(3)-3\]
Now I also asked you to replace the b with (-6a) like this
\[f(3)=9a+(-6a)(3)-3\]

Answer 13

now remember f(3)=15 so we have:
\[9a+(-6a)(3)-3=15 \]

Answer 14

Do you think you can solve this for a?

Answer 15

no, i'm so confused .. I'm so sorry, this is literally the 60th math question i'm finishing up today.

Answer 16

Can you tell me what confuses you? Like just the whole problem?

Answer 17

the whole problem but i'll ask my teacher to give me a hand on monday.

Answer 18

So you don't have any clue that the vertex of \[y=ax^2+bx+c \text{ is } (\frac{-b}{2a}, f(-\frac{b}{2a}))?\]

Answer 19

Like do I need to convince you why that is?
Or is it something else?

Answer 20

no clue :(

Answer 21

What? I'm confused. What do you mean? No clue what I'm asking? Or you don't know that is the vertex formula for anything written in this form y=ax^2+bx+c where a does not equal 0?

Answer 22

oh the vertex , what should I do to get it?

Answer 23

You already have the vertex.
The vertex is (3,15).

Answer 24

But we know that \[(3,15)=(\frac{-b}{2a}, f(\frac{-b}{2a}))\]

Answer 25

since that is the vertex formula for y=ax^2+bx+c where a does not equal 0
and the vertex was actually given (3,15)

Answer 26

So we want both of those x-coordinates to be equal
and we want both of those y-coordinates to be equal

Answer 27

oh yes, i meant the values of a and b

Answer 28

\[3=\frac{-b}{2a} \text{ <-the x-'s } \text{ the y's -> } 15=f(-\frac{b}{2a})\]
Or we could just say \[15=f(3) \text{ since } f(\frac{-b}{2a})=f(3) \]

Answer 29

f(3)=15 means we have
\[f(3)=a(3)^2+b(3)-3=15\]
\[a(3)^2+b(3)-3=15 \]

Answer 30

now remember you also have \[3=\frac{-b}{2a}\]
Solve this for b.

Answer 31

I'm trying to get you to get b by itself.
b is being divided by (-2a) so multiply both sides by (-2a)

Answer 32

b = -6a

Answer 33

Right. now go to the equation I posted that had the other a and b and replace that b with (-6a)

Answer 34

The equation right above the one I asked you to solve for b.

Answer 35

okay

Answer 36

so what do i do?

Answer 37

I'm still waiting for you to show me the rewrite of equation I wrote above the one you solved for b.
You know I asked you to replace that b with (-6a) since b equals (-6a)

Answer 38

Remember we have also \[a(3)^2+b(3)-3=15 \]

Answer 39

I'm trying to get you to replace that b with (-6a)
just put (-6a) instead of b that is all I'm asking.