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

#1**+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

#1**+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

#3**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