How do you find the equation of the parabola with vertex (-3, 1) and passing thru (-5, -11) if its axis of symmetry is parallel to the Y-axis?

Question

Isaiah Anthony · Accepted Answer

#y=-3x^2-18x-26# #"the equation of a parabola in "color(blue)"vertex form"# is. #color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))# #"where "(h,k)" are the coordinates of the vertex and a is"# #"a multiplier"# #"here "(h,k)=(-3,1)# #rArry=a(x+3)^2+1# #"to find a substitute "(-5,-11)" into the equation"# #-11=4a+1rArr4a=-12rArra=-3# #rArry=-3(x+3)^2+1larrcolor(red)" in vertex form"# #"distributing and simplifying gives"# #y=-3x^2-18x-26larrcolor(red)"in standard form"# graph{-3x^2-18x-26 [-11.25, 11.25, -5.625, 5.625]}

Ethan Adkins · Answer

undefined To find the equation of the parabola, you can use the standard form equation for a parabola with a vertical axis of symmetry:

$$ (x - h)^2 = 4p(y - k) $$

Where (h, k) is the vertex of the parabola and $ p $ is the distance from the vertex to the focus (or from the vertex to the directrix).

Given that the vertex is (-3, 1) and a point on the parabola is (-5, -11), you can use these values to find $ p $:

$$ (x - (-3))^2 = 4p(y - 1) $$
$$ (x + 3)^2 = 4p(y - 1) $$

Now, substitute the coordinates of the point (-5, -11):

$$ (-5 + 3)^2 = 4p(-11 - 1) $$
$$ (-2)^2 = 4p(-12) $$
$$ 4 = -48p $$
$$ p = -\frac{1}{12} $$

Now, substitute $ p $ back into the equation:

$$ (x + 3)^2 = 4\left(-\frac{1}{12}\right)(y - 1) $$
$$ (x + 3)^2 = -\frac{1}{3}(y - 1) $$

Therefore, the equation of the parabola is:

$$ (x + 3)^2 = -\frac{1}{3}(y - 1) $$