Question 1033359: Write the equation of the parabola y = x² - 6x - 29 in standard form.
A.y + 29 = (x - 3)²
B.y + 38 = (x - 3)²
C.y + 29 = (x - 2)²
D.y + 33 = (x - 2)²
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Write the equation of the parabola y = x² - 6x - 29 in standard form.
y + 29 + ? = x^2-6x + ?
y + 29 + 9 = x^2-6x+3^2
y + 38 = (x-3)^2
----------------
Cheers,
Stan H.
-------------
A.y + 29 = (x - 3)²
B.y + 38 = (x - 3)²
C.y + 29 = (x - 2)²
D.y + 33 = (x - 2)²
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! to convert from standard form to vertex form, you are completing the square on the standard form of the equation.
start with y = x^2 -6x - 29
add 29 to both sides of the equation to get y + 29 = x^2 - 6x
complete the square on the right side of the equation to get y + 29 = (x-3)^2 - 9
subtract 29 from both sides of the equation to get y = (x-3)^2 - 38
that's the vertex form of the equation.
the general form of the vertex form of the equaiton is y = a * (x-h)^2 + k.
in your equation, a = 1, h = 3, k = -38.
the vertex is (h,k) which is equal to (3,-38).
your equation is y = (x-3)^2 - 38.
if you add 38 to both sides of the equation, you will get y + 38 = (x-3)^2.
that matches selection B.
the graph shown below shows you that the equations of y = x^2 - 6x - 29 and y = (x-3)^2 - 38 are equivalent because they generate the identical graph.
they also show you that the vertex is at (3,-38), as provided by the vertex form of the equation.
|
|
|
