GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
For the parabola in the xy-plane, find the following. 2y =x −
[#permalink]
24 May 2019, 08:15
24 May 2019, 08:15
1
Expert Reply
10
Bookmarks
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) The x-intercepts
(b) The y-intercept
(c) Coordinates of the vertex
Show: :: OA
(a) and x=−2 x=6 (b) y=−12 (c) (2,−16)
Math Review
Question: 19
Page: 245
Difficulty: medium
Question: 19
Page: 245
Difficulty: medium
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: For the parabola in the xy-plane, find the following. 2y =x −
[#permalink]
28 May 2019, 06:26
28 May 2019, 06:26
3
1
4
Bookmarks
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) The x-intercepts
(b) The y-intercept
(c) Coordinates of the vertex
Show: :: OA
(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 vertex
Take: 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
General Discussion
Manager
Joined: 10 Dec 2020
Posts: 50
Given Kudos: 955
GRE 1: Q166 V162
For the parabola in the xy-plane, find the following. 2y =x −
[#permalink]
18 Dec 2021, 13:26
18 Dec 2021, 13:26
for a quad eqn: \(y = ax^2 + bx + x\)
vertex of parabola ------> (\(\frac{-b}{2a}\), k) ref
k is y's value in eqn when \(x = \frac{-b}{2a}\)
vertex of parabola ------> (\(\frac{-b}{2a}\), k) ref
k is y's value in eqn when \(x = \frac{-b}{2a}\)
Therefore, for the mentioned parabola \(y = x^2 −4x−12\):
\(\frac{-b}{2a} = \frac{-(-4)}{(2*1)} = 2\)
\(k = 2^2 - (4*2) - 12 = -16\)
Coordinates of vertex = (2, -16)
