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# Quadratic Functions derived from vertex and a point

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

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

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..

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

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)

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

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
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Thank you Melody!

Aug 28, 2014
#3
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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.

Aug 28, 2014
#4
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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

Aug 28, 2014