calculusxy
  • calculusxy
Help with finding the vertex of parabolas!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
calculusxy
  • calculusxy
\[\large \frac{1}{2}(y + 4) = (x - 7)^2 \]
calculusxy
  • calculusxy
@Nnesha
WolframWizard
  • WolframWizard
Expand the equation first and solve for y to get the equation in standard form.

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calculusxy
  • calculusxy
\[\frac{1}{2}y + 2 = x^2 - 14x + 49\]
Vocaloid
  • Vocaloid
actually, you don't need to rewrite the equation at all
Nnesha
  • Nnesha
the standard form is (x-h)^2=4p(y-k) where (h,k) is the vertex
Nnesha
  • Nnesha
well that's the vertex form you will see that n conics unit but in algebra we use y=a(x-h)^2+k as vertex form
calculusxy
  • calculusxy
i already have (x - h)^2 + with (x-7)^2. how would i find out the value of k?
Nnesha
  • Nnesha
what number is replaced by h and k ??
calculusxy
  • calculusxy
h = 7?
calculusxy
  • calculusxy
k = 2 ?
Nnesha
  • Nnesha
\[\rm \large \frac{1}{2}(y + 4) = (x - 7)^2 \] compare this with the equation ive posted above
Nnesha
  • Nnesha
k=2 ???
calculusxy
  • calculusxy
i don't understand
Nnesha
  • Nnesha
here is an example \[\large\rm (x-h)^2=4p(y-k)\] where(h,k) is the vertex remember both h and k are positive here is an example \[\rm (x-\color{ReD}{5})^2=4(y+\color{blue }{3})\] where (5 ,-3) is the vertex
Nnesha
  • Nnesha
do you know how to find h and k value from this equation \[\rm y=3(x-4)^2+5\] ?
calculusxy
  • calculusxy
would the h = 4 and k = 5?
Nnesha
  • Nnesha
yes right that's how you have to find h and k value from this equation \[\rm (x-h)^2 =4p (y-k)\]
calculusxy
  • calculusxy
what does the p do in 4p?
idku
  • idku
if you have the equation of the parabola you just need to somplete the square
idku
  • idku
if you know the completing the square method, then you are ready tackle these.
idku
  • idku
Example: Find the vertex, y=x²+4x-3 y=x²+4x+4-4-3 Notice that (x²+4x+4) is a "perfect square trinomial" (you have to know what this term means). y=(x²+4x+4)-7 y=(x+2)²-7 vertex is (2,7) (and since parabola opens up the vertex is the minimum point. the minimum is -7, and it occurs at x=2)
idku
  • idku
I made an error
Nnesha
  • Nnesha
4p is same as `a` in the other equation if this is not conics section then i guess you should expand it first
idku
  • idku
the vertex in my example should be (-2,-7)
Nnesha
  • Nnesha
unit*
calculusxy
  • calculusxy
i have not heard of conics. so i think that i will stick with the basic a(x - h)^2 + k
Nnesha
  • Nnesha
alright then you should expand it
calculusxy
  • calculusxy
so i got: \[\frac{1}{2}y + 2 = x^2 - 14x + 49\]
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @calculusxy \[\frac{1}{2}y + 2 = x^2 - 14x + 49\] \(\color{blue}{\text{End of Quote}}\) that's correct solve for y
calculusxy
  • calculusxy
1/2y + 2 = x^2 - 14x + 49 1/2y = x^2 - 14x + 47 y = 2x^2 - 28x + 94
Nnesha
  • Nnesha
now two ways to find the vertex you can complete the square or use -b/2a formula up to you :=))
calculusxy
  • calculusxy
-(-28)/4 = 7 2(49) - 28(7) + 94 98 - 196 + 94 -4 (7, -4)
calculusxy
  • calculusxy
2(x - 7)^2 - 4
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\)
calculusxy
  • calculusxy
thanks!

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