Find the equation of parabola whose axis of symmetry is x=-3 and passes through points (-1,3) and (2,-5)?

Jul 24, 2017

Answer:

Please see below.

Explanation:

Given that the axis is $x = - 3$, we know the vertex is at $x = - 3$, so the equation can be written in the form

$y = a {\left(x + 3\right)}^{2} + k$

Now use the two points given to get:

From $\left(- 1 , 3\right)$, we have $3 = a {\left(2\right)}^{2} + k$ so $4 a + k = 3$ and $k = 3 - 4 a$

From $\left(2 , - 5\right)$, we get $- 5 = a {\left(5\right)}^{2} + k$ so $25 a + k = - 5$

Replace $k$ in the last equation with $3 - 4 a$ to get

$25 a + \left(3 - 4 a\right) = - 5$, so

$21 a = - 8$ and $a = - \frac{8}{21}$

Using $k = 3 - 4 a$ again we get $k = 3 - \left(- \frac{32}{21}\right) = \frac{63}{21} + \frac{32}{21} = \frac{95}{21}$.

The equation is $y = - \frac{8}{21} {\left(x + 3\right)}^{2} + \frac{95}{21}$

If course, there are other ways to write the answer.

graph{(y+8/21(x+3)^2-95/21)(x+3)((x+1)^2+(y-3)^2-0.02)((x-2)^2+(y+5)^2-0.02)=0 [-12, 12, -6, 6]}