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1. v(-4,3); containing (-6,11)

2. containing (0,6) and (3,-1)

My answer:

1. v(-4,3); pt (-6,11)

y=a(x-h)^2+k

11= a(-6-(-3)+(-3)

11=a(-6+3)-3

11= a(-3)-3

11=-3a-3

11+3=-3a

-14/3=a

or -4 2/3

 

Vertex form: y= -4 2/3 (x+4)^2 -3

Standard form: I don't know how to get it... Please tell me how..

 

2. I don't know how to solve this one... Please tell me how to solve this, get the vertex form and the standard form...

 

Please correct any error that I made i my solution.. I am not that good at math..

Phanolo  Aug 28, 2014

Best Answer 

 #1
avatar+91451 
+5

 

V(-4,3) point (-6,11)

There is more than one parabola that fits this description - an infinite number I think.  You need 3 points for a unique parabola.  However if the axis of symmetry is parallel to the y axis then there is only one and this is a fair assumption for your question.

 

Let's look at your answer:

1. v(-4,3); pt (-6,11)

y=a(x-h)^2+k          alright

y=a(x--4)^2+3         added

y=a(x+4)^2+3          added

The h and k come directly from the vertex.  You sub those in first and then use the other point to determine the value of a. 

11= a(-6-(-3)+(-3)     incorrect  (I won't worry about the rest)

I'll redo this for you (your method)

 

$$v(-4,3); pt (-6,11)\\
\begin{array}{rlll}
\mbox{First sub in the vertex}\\
y&=&a(x-h)^2+k \\
y&=&a(x--4)^2+3 \\
y&=&a(x+4)^2+3 \\
\mbox{Now sub in the other point to find a}\\
11&=&a(-6+4)^2+3 \\
8&=&a(-2)^2 \\
8&=&4a \\
a&=&2 \\
\mbox{Therefore}\\
y&=&2(x+4)^2+3 &\mbox{This is the vertex form}\\
\end{array}$$

 

$$\mbox{Now I will convert it to standard form }\\\\
\begin{array}{rlll}
(x+4)^2&=&\frac{1}{2} (y-3)\\
x^2+8x+16&=&\frac{1}{2} (y-3)\\
2x^2+16x+32&=& (y-3)\\
2x^2+16x+35&=& y\\
y&=& 2x^2+16x+35\\
\end{array}\\\\$$

 

(I have edited this to get rid of irrelevant calculations)

Melody  Aug 28, 2014
Sort: 

4+0 Answers

 #1
avatar+91451 
+5
Best Answer

 

V(-4,3) point (-6,11)

There is more than one parabola that fits this description - an infinite number I think.  You need 3 points for a unique parabola.  However if the axis of symmetry is parallel to the y axis then there is only one and this is a fair assumption for your question.

 

Let's look at your answer:

1. v(-4,3); pt (-6,11)

y=a(x-h)^2+k          alright

y=a(x--4)^2+3         added

y=a(x+4)^2+3          added

The h and k come directly from the vertex.  You sub those in first and then use the other point to determine the value of a. 

11= a(-6-(-3)+(-3)     incorrect  (I won't worry about the rest)

I'll redo this for you (your method)

 

$$v(-4,3); pt (-6,11)\\
\begin{array}{rlll}
\mbox{First sub in the vertex}\\
y&=&a(x-h)^2+k \\
y&=&a(x--4)^2+3 \\
y&=&a(x+4)^2+3 \\
\mbox{Now sub in the other point to find a}\\
11&=&a(-6+4)^2+3 \\
8&=&a(-2)^2 \\
8&=&4a \\
a&=&2 \\
\mbox{Therefore}\\
y&=&2(x+4)^2+3 &\mbox{This is the vertex form}\\
\end{array}$$

 

$$\mbox{Now I will convert it to standard form }\\\\
\begin{array}{rlll}
(x+4)^2&=&\frac{1}{2} (y-3)\\
x^2+8x+16&=&\frac{1}{2} (y-3)\\
2x^2+16x+32&=& (y-3)\\
2x^2+16x+35&=& y\\
y&=& 2x^2+16x+35\\
\end{array}\\\\$$

 

(I have edited this to get rid of irrelevant calculations)

Melody  Aug 28, 2014
 #2
avatar+44 
0

Thank you Melody!

Phanolo  Aug 28, 2014
 #3
avatar+91451 
0

2)  This second one looks tougher.

I assume you would use simultaneous equations.  Lets give that a go.  

containing (0,6) and (3,-1)

$$\\y=a(x-h)^2+k\\\\
6=a(0-h)^2+k\\
6-k=ah^2 \qquad (eqn\;1)\\\\
-1=a(3-h)^2+k\\
-1=a(9+h^2-6h)+k\\
-1-k=ah^2-6ah+9a \qquad(eqn\:2)\\\\
6-k=ah^2 \qquad (eqn\;1)\\
-1-k=ah^2-6ah+9a \qquad(eqn\:2)\\
eqn1-eqn2\\
6-k-(-1-k)=ah^2-(ah^2-6ah+9a)\\
7=6ah-9a\\
7=a(6h-9)\\
a=\dfrac{7}{6h-9}\\\\$$

 

I've been writing this and putting it into LaTex at the same time - there could easily be errors.

anyway.  I am assuming that from here you can go back sub this value of a into the simultaneous equations and come up with a value for h and k.  

Why don't you give it a go and see if you can finish it?   I'll look at it some more if you get stuck.

Melody  Aug 28, 2014
 #4
avatar+91451 
0

The second question is really puzzling me and I would like more input from other mathematicians.

I have made this graph and it appears to be telling me that one approx answer is h=4.2 and k=-1.6

but it is confusing.  Would anyone like to comment.

 

 

https://www.desmos.com/calculator/0wieewtzra

Melody  Aug 28, 2014

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