Question

Without drawing the graph of the equation, determine how many points the parabola has in common with the x-axis and whether its vertex lies above, on, or below the x-axis.

y = 4x2 – 12x + 12

Answer 1

A.2 points in common; vertex above x-axis
B.no points in common; vertex above x-axis
C.1 point in common; vertex on x-axis
D.2 points in common; vertex below x-axis

Answer 2

@jim_thompson5910

Answer 3

@Loser66

Answer 4

"determine how many points the parabola has in common with the x-axis"

basically, they want the x-intercepts for this bit

Answer 5

So I should use slope intercept form?

Answer 6

this is a quadratic, not a linear equation

Answer 7

How do I do that?

Answer 8

^2?

Answer 9

you use the quadratic formula to solve for x in 4x^2 – 12x + 12 = 0

Answer 10

set each factor equal to zero, and then solve each factor?

Answer 11

In that case is this right?
4x^2+12=0
-12x+12=0

Answer 12

I'm not sure why you broke that up into two equations

Answer 13

How else should I do it?

Answer 14

use the quadratic formula

Answer 15

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 16

in this case,

a = 4, b = -12 c = 12

Answer 17

... Wow that looks scary

Answer 18

hopefully you have seen it before

Answer 19

I will try

Answer 20

x=12+-Square root -12^2-4*4*12/8

Answer 21

\[x=12+-\sqrt{-144-192/8}\]

Answer 22

Right? @jim_thompson5910

Answer 23

what is b^2-4ac equal to?

Answer 24

This? x=12+−square root 144−192/8

Answer 25

no just b^2-4ac

Answer 26

evaluate b^2-4ac when a = 4, b = -12 c = 12

Answer 27

I tried! I got x=12+-Square root -12^2-4*4*12/8 :(

Answer 28

b^2-4ac

(-12)^2-4(4)(12)

144 - 192

-48

that's what I was asking for

Answer 29

So was I partially correct?

Answer 30

so we would then have

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-(-12) \pm \sqrt{(-12)^2-4(4)(12)}}{2(4)}\]

\[\Large x=\frac{12 \pm \sqrt{-48}}{8}\]

Answer 31

I got -12^2-4*4*12 originally

Answer 32

How will that help me find the answer though @jim_thompson5910

Answer 33

that discriminant is negative

what does that tell you?

Answer 34

D. Below?

Answer 35

what does the discriminant determine in general?

Answer 36

I don't know..

Answer 37

have a look at this page

http://www.mathwarehouse.com/quadratic/discriminant-in-quadratic-equation.php

Answer 38

Oh! It tells If the solution is a real number or an imaginary number, If the solution is rational or if it is irrational, and If the solution is 1 unique number or two different numbers

Answer 39

good

Answer 40

so the discriminant of -48 tells us what?

Answer 41

That it is irrational and there is No Real Solutions

Answer 42

and theres Two Imaginary Solutions

Answer 43

well if you have irrational solutions, then you have a solution

so you can't have "no real solutions" and have them be irrational at the same time

Answer 44

but you saying " theres Two Imaginary Solutions " is correct

Answer 45

what does this mean for the x-intercepts?

Answer 46

They will be negative? or below?

Answer 47

a solution to an equation like 4x^2 – 12x + 12 = 0 visually corresponds to the x-intercept

this is assuming you get a real numbered solution

Answer 48

but the solutions are imaginary numbers (not real numbers)

Answer 49

2 points in common?

Answer 50

but I didn't get a real numbered solution :c

Answer 51

so are there 2 x-intercepts?

Answer 52

Yes, that is what I think

Answer 53

but the discriminant is negative

Answer 54

negative discriminant ----> no real solutions ----> ???

Answer 55

No points in common?

Answer 56

don't worry about the answer choices right now

Answer 57

Ok

Answer 58

if we have no real solutions, then how many x-intercepts are there?

Answer 59

use that page I posted

Answer 60

So it means 2 imaginary solutions.. thats why I thought 2 x intercepts

Answer 61

I am using it still, Thanks for the page.

Answer 62

you'd have 2 x-intercepts if you had 2 real solutions

Answer 63

another way to think of it as a general rule

the number of unique real solutions tells you how many unique x-intercepts you have

Answer 64

So 2 imaginary is no x intercepts?

Answer 65

yep

Answer 66

there are no x-intercepts

Answer 67

so the graph doesn't cross the x axis

Answer 68

but.. all my choices have something to do with the vertex on X

Answer 69

Above, Below, or On... But you said it doesn't cross

Answer 70

I'm not talking about the vertex at the moment

Answer 71

I'm just referring to the x-intercepts

Answer 72

Oh, umm so the descriminate is -48 and there is no X intercepts

Answer 73

remember they said "determine how many points the parabola has in common with the x-axis"

so they're saying "find the number of x-intercepts"

Answer 74

But there is none

Answer 75

I am sorry... but I don't get how this will help me get points or a vertex :(

Answer 76

Intercept has nothing to do with vertex right?

Answer 77

so having no x-intercepts is their way of saying there are no points that are in common with the x axis

Answer 78

Ok, and the vertex is above the X axis?

Answer 79

based on the answer choices, it has to be

but it's always a good idea to know how to find the vertex

Answer 80

this page gives a good example

http://www.wikihow.com/Find-the-Vertex-of-a-Quadratic-Equation

Answer 81

I posted this question
an hour ago

Thanks for helping me through it c:

Answer 82

you're welcome