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**+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

#2**+8 **

Best Answer

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