+44 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..
+118609
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)
+118609
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)
+118609 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.
