What is the equation of the parabola that has a vertex at # (-4, 16) # and passes through point # (0,0) #?

1 Answer
Feb 9, 2016

Let us solve this problem by substituing both points into a parabola equation: #ax^2 + b x + c = y(x)#

Explanation:

  • First of all, let us substitute #(0,0)#:

#ax^2+bx+c= y(x) rightarrow a cdot 0^2 + b cdot 0 + c = y(0) rightarrow c = 0#

Thus, we obtain the independent term in equation, getting #ax^2 + bx = y(x)#.

  • Now, let us substitute the vertex, #(-4, 16)#. We get:

#a cdot (-4)^2 + b cdot (-4) = 16 rightarrow 16 a - 4 b = 16 rightarrow 4 a - b = 4#

Now, we have a relation between #a# and #b#, but we cannot determine them uniquely. We need a third condition.

  • For any parabola, the vertex can be obtained by:

#x_"vertex" = {-b}/{2a}#

In our case:

#x_"vertex" = -4 = {-b}/{2a} rightarrow b = 8 a#

  • Finally, we must solve the system given by:

#{4a-b=4 ; b = 8a}#

Replacing #b# from second equation to the first one:

#4a-(8a)=4 rightarrow -4 a = 4 rightarrow a = -1#

And finally:

#b = -8#

This way, the parabola equation is:

#y(x) = -x^2 - 8x#