Question

Is there a x-intercept?
y - 2 = 2(x+1)^2

Answer 1

Here is a graph, it doesnt look like there is one

Answer 2

There will be imaginary roots.

Answer 3

Let's check is there any x for which is y = 0

$$ y = 2(x+1)^2 + 2 = 2((x + 1)^2 + 1) $$
$$ y = 0 \rightarrow (x+1)^2 = -1 $$ which is impossible in set of real numbers

Answer 4

so it becomes an imaginary number?
@fantastic001

Answer 5

I need to be able to find the:
Vertex: (-1, 2)
x-intercept: not sure
y-intercept: 4
axis o symmetry: -1

Answer 6

@feegee if x is in set of comples numbers, then it is complex number, in real numbers, x such that y = 0 does not exist because in real numbers square is always positive

Answer 7

@fantastic001 okay, so from everything I've gotten from you so far- well my understanding, is that there is no x-axis, it doesnt exist.

Answer 8

I'm sorry, I meant x-intercept**

Answer 9

yeah, it doesn't exist x-axis intersection because y is never 0 :)

Answer 10

alright! Do you mind if I bring up another question which is a bit trickier than? @fantastic001

Answer 11

i'll try to help if i can :)

Answer 12

@fantastic001 great!

It's the same thing, where I need to find the:
vertex:
x-intercept:
y-intercept: -6
axis of symmetry: 1

Answer 13

okay, so looks like the vertex is : (1, -6.5)

Answer 14

ok, this is a parabola and equation for parabola is $$ y = ax^2 + bx + c $$ in y-intersection $$ x = 0 $$
so we have $$ y_{intersect} = a*0^2 + b*0 + c = c $$ and you say that $$ y_{intersect} = -6 $$ so $$ c = -6 $$

ok we have $$ c = -6 $$ , we need a and b right?

look at graph you have point (1, -6.5) , put that in equation

$$ -6.5 = a*1^2 + b -6 $$
$$ 0.5 = a +b $$
OK, , let's put another information, look at graph again, you have point (2, -6) so we have

$$ -6 = 4a + 2b - 6 $$

$$ 0 = 4a + 2b $$
$$ 0.5 = a+b $$
it is system of linear equations, we know how to solve it :)

$$ b = -2a $$
$$ a = -0.5 $$
$$ b = 1 $$

now make a complete equation

$$ y = -0.5*x^2 + x - 6 $$

now solve $$ y = 0 $$ and you'll get intersections with x-axis (there are two intersections !)

Answer 15

okay, correct me if I'm wrong but I'm getting; \[1 + i \sqrt{11} or 1 - i \sqrt{11}\]

Answer 16

@fantastic001

Answer 17

i think you're right

Answer 18

hmm but hraph on a picture and graph of $$ y = -0.5x^2 + x - 6 $$ are not same hmm

Answer 19

The function for the graph is in vertex form: y - 2 = 2 (x+1)^2

Answer 20

@fantastic001

Answer 21

ok, to find y-intersect just solve $$ y = 2(x + 1)^2 + 2 $$ at $$ x = 0 $$ so $$ y = 4 $$
but this equation has no solution for $$ y = 0 $$ in set of real numbers

Answer 22

looks like neither of my problems have a solution for the x-intercept.

Answer 23

for $$ y = 0 $$ you have $$ x = -1 \pm i $$ but this is complex number

Answer 24

yeah, looks like they both are complex numbers

Answer 25

for both x-intercepts