Question

Need help on a few problems.. Some are mostly done. Quadratic Function

Answer 1

I know its a positive graph.
Vertex- (-3,0)
y-intercept- 9

Answer 2

vertex form of quadratic equation is y = a(x-h)^2 +k
where (h,k) is the vertex point so just sub h and k for (-3,0 )

Answer 3

you can pick any (x,y) on the parabola graph to find a
and yes right it would be positive bec graph have the min point

Answer 4

Okay? That doesnt help me finish solving this.

Answer 5

0= (-3)^2 +(-6a)(-3) +9

Answer 6

a(-3)^2 ***

Answer 7

\[\huge\rm y=a(x-\color{ReD}{h})^2+\color{blue}{k}\]
i don't know how u got 6a first substitute (h,k) for (-3,0) w

Answer 8

My example does (- b over 2a, f(- b over 2a)

- b over 2a=-3
b=-6a

Answer 9

f(x)= ax^2 +bx+c

Answer 10

that's the formula to find vertex but you already got the vertex by just looking at the graph

Answer 11

ahh i see so you're trying each options to get (-3,0) vertex ? backwards

Answer 12

you can do that too
\[( \frac{-b}{2a} , f(\frac{-b}{2a})\]
try each option sub a and b for their values to find (-3,0) vertex point

Answer 13

... What? I'm just following the example.. This is my first attempt at this problem. My example does a negative graph.

Answer 14

After finding -6 and 0 it says to plug everything into the equation

Answer 15

\(\color{blue}{\text{Originally Posted by}}\) @Destinyyyy
0= (-3)^2 +(-6a)(-3) +9
\(\color{blue}{\text{End of Quote}}\)
i'm sorry i just don't understand this equation.

Answer 16

can i see the example ?

Answer 17

f(x)= ax^2 +bx +c

-3 is x and h
-6 is b
9 is c

f(x)= a(-3)^2 +(-6a)(-3)+9
replace f(x) with 0

Answer 18

If I try to screenshot the example it will be in multiple pictures..

Answer 19

oky then nvm

Answer 20

i see.
so b is just -6 not -6a

Answer 21

I got a= -.3

Answer 22

\[0=a(-3)^2+(-6)(-3)+9\] so it should be like his
but why are we using -6 and9 ?

Answer 23

i know two ways to find equation from a graph
use the vertex form plug in vertex point
and doing backwards
try each option to get the the same vertex that's on the graph :=))

Answer 24

My example keeps the a with -6 .. So the two a's are added together. Then divided. I find a.. Then it plugs a into the equation and thats my answer.

Answer 25

b= -6a and c= 9 Thats why they are being used.

Answer 26

\[0=a(-3)^2+(-6a)(-3)+9\] that will give u a = -1/3 which is not correct

Answer 27

for this question
we don't know what a b and c are we have to find a b and c values

Answer 28

The answer is x^2 +6x+9

But a is coming out wrong. It should be a positive 1.

Answer 29

yes right that's the answer let's do it \[\huge\rm y=a(x-h)^2+k\]
(h,k) --->(-3,0)
\[\huge\rm y=a(x-(-3))^2+0\]
\[\huge\rm y=a(x+3)^2\]
now pick any two points on the parabola graph to find value of a

Answer 30

0+ anything = anything so that's how it's a(x+3)^2

Answer 31

Yeah I know how to do it that way. I did the first 3 that way. But this one said to do it different

Answer 32

ohh okay
then i would like to see the example

Answer 33

Okay .. One second

Answer 34

Thats all of it.

Answer 35

ahh i see so first we should use the formula \[\huge\rm \frac{ -b }{ 2a } = -3\]
x-coordinate of the vertex is -3 so set it equal to -3
and then solve for b

Answer 36

-b = 2a * -3
b= ??
remember negative divide by negative = positive

Answer 37

Thats where I messed up.

Answer 38

I missed the front negative.. I have a few more. Could you help me with them?

Answer 39

yes right b =6a not -6a \[0=a(-3)^2+(6a)(-3)+9\]

Answer 40

sorry i have to go now
later! :=))