The graph of the parabola defined by the equation y=(x-2)^2 + 3 is rotated 180 degrees about its vertex, then shifted 3 units to the left, then shifted 2 units down. The resulting parabola has zeros at x=a and x=b. What is a+b?
Help me solve this please! Thanks
y = (x - 2)^2 + 3
This parabola has its vertex at (2, 3) . Rotating it 180° about its vertex has the same effect as flipping it over the x-axis and then shifting it up 6 units. So our new parabola has the equation...
y = -[ (x - 2)^2 + 3 ] + 6
Now let's shift it 3 units to the left...
y = -[ (x - 2 + 3)^2 + 3 ] + 6
And 2 units down...
y = -[ (x - 2 + 3)^2 + 3 ] + 6 - 2 Then simplify.
y = -(x +1)^2 + 1
The zeros of this parabola are the x values when...
0 = -(x +1)^2 + 1
(x +1)^2 = 1
x + 1 = ±√1
x + 1 = 1 or x + 1 = -1
x = 0 or x = -2
And the sum of the zeros = 0 + -2 = -2
Here's a graph of the original parabola and the transformed parabola.
y = (x - 2)^2 + 3
If this is rotated 180° about its vertex, the vertex does not change..........the new function is
y = - (x- 2)^2 + 3
Shifting this to the left 3 units left and 2 units down results in the function
y = - (x + 1)^2 + 1
To find the zeroes, we have
-(x + 1)^2 + 1 = 0
-(x +1)^2 = -1
(x + 1)^2 = 1
x + 1 = ±1
So
x + 1 = 1 ⇒ x = 0
x + 1 = - 1 ⇒ x = -2
So a + b = 0 + -2 = - 2
Here's the graph : https://www.desmos.com/calculator/zbkwtqadyr