Carcass wrote:
For the parabola in the xy-plane, ﬁnd the following. \(y =x^2 −4x−12\)
(a) The x-intercepts
(b) The y-intercept
(c) Coordinates of the vertex
(a) and x=−2 x=6 (b) y=−12 (c) (2,−16)
(a) The x-intercepts The x-intercepts are points where the y-coordinate is ZERO.
So, replace y with 0 to get: \(0=x^2−4x−12\)
Factor to get: \(0=(x-6)(x+2)\)
So, either x = 6 or x = -2
So, the
x-intercepts are 6 and -2(b) The y-intercept The y-intercept is the point where the x-coordinate is ZERO.
So, replace x with 0 to get: \(y=0^2−4(0)−12\)
Evaluate: \(y=−12\)
So, the
y-intercept is -12(c) Coordinates of the vertexTake: y = x² − 4x − 12
Rewrite as: y = x² − 4x
+ 4 − 12
- 4 [the net effect of adding 4 and subtracting 4 is ZERO]Rewrite as: y = (x² − 4x
+ 4) − 16
Factor to get: y = (x - 2)² − 16
When we compare this to the graph of the BASE equation y = x², we can see that it has been shifted 2 units to the RIGHT, and 16 units DOWN.
So, the
vertex is now at (2, -16)Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep