How to find the equation of a Parabola with vertex (0,-9) and passing through (6,-8)?

Question

Eli Assouline · Accepted Answer

#y = x^2/36 -9# #color(white)("XXXX")##y = m(x-a)^2+b# #color(white)("XXXX")##color(white)("XXXX")#, where the vertex is at #(a,b)#, is the general vertex form for a parabola. Since the desired parabola's vertex is located at #(0,-9)#, the equation becomes: #color(white)("XXXX")##y = m(x-0)^2-9#. Moreover, #color(white)("XXXX")##-8 = m(6-0)^2-9# is a solution point on this parabola, since #(x,y) = (6,-8)#. ##1 = 36m# #color(white)("XXXX") ##m = 1/36# #color(white)("XXXX") Hence, either #color(white)("XXXX")##y = x^2/36 -9# or #color(white)("XXXX")##y = 1/36(x-0)^2-9#.

Nathan Axel · Answer

undefined The equation of the parabola is y = (1/6)x^2 - 9.

Julia Adams · Answer

undefined To find the equation of a parabola given its vertex and a point it passes through, we can use the standard form of the equation for a parabola with vertex at the origin ($h = 0, k = -9$):

$$ y = a x^2 - 9 $$

We substitute the coordinates of the given point (6, -8) into this equation and solve for $a$:

$$ -8 = a \cdot 6^2 - 9 $$

$$ -8 = 36a - 9 $$

$$ 36a = -8 + 9 $$

$$ 36a = 1 $$

$$ a = \frac{1}{36} $$

Now we have the value of $a$, we can write the equation of the parabola:

$$ y = \frac{1}{36} x^2 - 9 $$

So, the equation of the parabola with vertex (0, -9) and passing through (6, -8) is:

$$ y = \frac{1}{36} x^2 - 9 $$