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.
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.
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.