Finding the y-intercept of a Parabola: A Hint

[Echo 6 Sierra]

I'm missing a step somewhere.

Without a calculator, graph y=3x^2-16x-12 by factoring and plotting zeros.

I have gotten as far as (x=-2/3) & (x=6) and know it's a parabola and pointing up because it has the positive x^2 so the graphing is easy enough, except...

The botb says the y-int is -100/3 but I can't find the bridge to get there in my notes or in the chapter.

a hint, please.

Thank you.

 

[Doc Al]

Originally posted by Echo 6 Sierra
... The botb says the y-int is -100/3 but I can't find the bridge to get there in my notes or in the chapter.

Must be a misprint. The y-intercept is the point where the curve intersects the y-axis, in other words: the value of y where x=0. That value is certainly not y = -100/3.

 

[Echo 6 Sierra]

Sorry, my error. The bottom coordinate of the parabola is
(8/3,-100/3), not the y-int.

Thank you for your prompt reply.

E6S

 

[HallsofIvy]

In other words, the vertex is at (8/3,-100/3).

You can find that by completing the square.
y=3x²-16x- 12= 3(x²- (16/3)x)- 12.
(16/3)/2= 8/3 and (8/3)²= 64/9
y= 3(x²- (16/3)x+ 64/9- 64/9)- 12
= 3(x²- (16/3)x+ 64/9)- 64/3- 12
= 3(x-8/3)²- 100/3

Now it is clear that when x= 8/3, y= -100/3. And that if x is any other number, then y= -100/3 plus something and so is higher. (8/3, -100/3) is the lowest point on the graph- the vertex.

 

[Echo 6 Sierra]

I'm vaaaaaguely familiar with completing the square. It was briefly touched on in my trig class.

I understand that if x is any other number <,> 8/3 that it will follow the parabola and make the y value change to a number greater than -100/3. Thanks.

This afternoon, one of the schools free tutors showed me to use -b/2a. Where does this little gem come from?

 

[HallsofIvy]

It's for those who prefer memorizing formulas rather than thinking!

If y= ax²+ bx+ c, then y= a(x²+ (b/a)x)+ c.

Now, think "(b/a)/2= b/(2a) and that squared is b²/4a²" so we need to add b²/4a² to complete the square:

y= a(x²+ (b/a)x+ b²/4a²-b²/4a²
= a(x²+ (b/a)x+ b²/4a²)- b²/4a+ c
= a(x+ b/(2a))²+ (c- b²/4a)

Now, we can see that, when x= -b/(2a), y= c- b²/4a. If x is any other number, the square is positive so y is larger. The vertex of the parabola (the lowest point) is at (-b/2a,c- b²/4a).

I think that completing the square is important enough that you should know how to do it, and practice it, without just memorizing that formula for the vertex.

 

[Echo 6 Sierra]

Thanks Halls. Exam today, this should come in handy.

E6S