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..
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)First sub in the vertexy=a(x−h)2+ky=a(x−−4)2+3y=a(x+4)2+3Now sub in the other point to find a11=a(−6+4)2+38=a(−2)28=4aa=2Thereforey=2(x+4)2+3This is the vertex form
Now I will convert it to standard form (x+4)2=12(y−3)x2+8x+16=12(y−3)2x2+16x+32=(y−3)2x2+16x+35=yy=2x2+16x+35
(I have edited this to get rid of irrelevant calculations)
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)First sub in the vertexy=a(x−h)2+ky=a(x−−4)2+3y=a(x+4)2+3Now sub in the other point to find a11=a(−6+4)2+38=a(−2)28=4aa=2Thereforey=2(x+4)2+3This is the vertex form
Now I will convert it to standard form (x+4)2=12(y−3)x2+8x+16=12(y−3)2x2+16x+32=(y−3)2x2+16x+35=yy=2x2+16x+35
(I have edited this to get rid of irrelevant calculations)
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+k6=a(0−h)2+k6−k=ah2(eqn1)−1=a(3−h)2+k−1=a(9+h2−6h)+k−1−k=ah2−6ah+9a(eqn2)6−k=ah2(eqn1)−1−k=ah2−6ah+9a(eqn2)eqn1−eqn26−k−(−1−k)=ah2−(ah2−6ah+9a)7=6ah−9a7=a(6h−9)a=76h−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.