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Parabola Algebra

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Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.

Guest May 5, 2014

#2
+94105
+8

Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.

Interesting question

(10,-14),     (1)

(-2,10)         (2)

Vertex(2, k)     (3)

$$(x-h)^2=4a(y-k)$$     where (h,k) is the vertex

$$(x-2)^2=4a(y-k)$$

Using (10,-14) we have $$64=4a(-14-k)$$

Using (-2,10) we have    $$16=4a(10-k)$$

$$\frac{64}{16}=\frac{4a(-14-k)}{4a(10-k)}\\\\ 4=\frac{-14-k}{10-k}\\\\ 4=\frac{-14-k}{10-k}\\\\ 40-4k=-14-k\\\\ 54=3k\\\\ k=18$$

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$$(x-2)^2=4a(y-18)$$

$$(10,-14) 64=4a(-32)\rightarrow -2=4a \rightarrow a=-0.5\\ check (-2,10) 16=4a(-8)\rightarrow a=-0.5\\$$

So the equation is

$$(x-2)^2=-2(y-18)$$

And that is that.  Can I have a thumbs up now please. OR if you don't understand ask for clarification.

Melody  May 6, 2014
#2
+94105
+8

Find the equation algebraically, of the parabola which passes through the points (10,-14) and (-2,10), and whose axis of symmetry is the equation x=2, using vertex form.

Interesting question

(10,-14),     (1)

(-2,10)         (2)

Vertex(2, k)     (3)

$$(x-h)^2=4a(y-k)$$     where (h,k) is the vertex

$$(x-2)^2=4a(y-k)$$

Using (10,-14) we have $$64=4a(-14-k)$$

Using (-2,10) we have    $$16=4a(10-k)$$

$$\frac{64}{16}=\frac{4a(-14-k)}{4a(10-k)}\\\\ 4=\frac{-14-k}{10-k}\\\\ 4=\frac{-14-k}{10-k}\\\\ 40-4k=-14-k\\\\ 54=3k\\\\ k=18$$

---------------------

$$(x-2)^2=4a(y-18)$$

$$(10,-14) 64=4a(-32)\rightarrow -2=4a \rightarrow a=-0.5\\ check (-2,10) 16=4a(-8)\rightarrow a=-0.5\\$$

So the equation is

$$(x-2)^2=-2(y-18)$$

And that is that.  Can I have a thumbs up now please. OR if you don't understand ask for clarification.

Melody  May 6, 2014